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Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat

Question 1.
Fill in the blanks and rewrite the sentences:
a. The amount of water vapour in air is determined in terms of its………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

b. If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their………….
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

c. When a liquid is getting converted into solid, the latent heat is……….  (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Question 2.
Observe the following graph. Considering the change in volume of water as its temperature is raised from 0 °C, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?

Answer:
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C. It is minimum at 4 °C. The volume of water goes on increasing in the range 4 °C to 10 °C.

In general, when a substance is heated, its volume goes on increasing with temperature. Thus, in the range 0 °C to 4 °C, behaviour of water is different from other substances. It is called anomalous behaviour of water.

Question 3.
What is meant by specific heat capacity?
How will you prove experimentally that different substances have different specific heat capacities?
Answer:
The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C is called the specific heat capacity of that object.

Question 4.
While deciding the unit for heat, which temperatures interval is chosen? why?
Answer:
While deciding the unit for heat, the temperature interval chosen is 14.5 °C to 15.5 °C. For the reason, see the information given in the following box.

Question 5.
Explain the following temperature vs time graph:

(Practice Activity Sheet – 1 and 4; March 2019)
Answer:
The graph shows what happens when a mixture of ice and water is heated continuously. The temperature of the mixture remains constant (0 °C) till all the ice melts as shown by the line AB. This temperature is the melting point of ice. On further heating, the temperature rises steadily from 0 °C to 100 °C as shown by the line BC, At 100 °C water starts converting into steam. This temperature is the boiling point of water. Further heating does not change the temperature and the conversion waters steam continues as shown by the line CD.

Question 6.
Explain the following:
a. the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contracting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature of the water at the surface continues to fall to 0 °C. Finally, the water at the surface is converted into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

b. How can you relate the formation of water droplets on the outer surface of a bottle taken out of a refrigerator with formation of dew?
Answer:
At a given temperature, there is a limit on how much water vapour the given volume of air can hold. The lower the temperature, the lower is the capacity of air to hold water vapour.

The temperature of a bottle kept in a refrigerator is lower than room temperature. Hence, when the bottle is taken out of the refrigerator, the temperature of the air surrounding the bottle is lowered. Therefore, the capacity of the air to hold water vapour becomes less. Hence, the excess water vapour condenses to form water droplets (like dew) on the outer surface of the bottle.

c. In cold regions in winter, the rocks crack due to anomalous expansion of water.
Answer:
Sometimes water enters into crevices of the rocks. When the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts a tremendous pressure on the rocks which crack and break up into small pieces.

Question 7.
a. What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

b. Which principle is used to measure the specific heat capacity of a substance?
Answer:
The principle of heat exchange is used to measure the specific heat capacity of a substance. This principle is as follows: If a system of two objects is isolated from the environment by keeping it inside a heat resistant box, then no energy can leave the box or enter the box. In this situation, heat energy lost by the hot object = heat energy gained by the cold object.

c. Explain the role of latent heat in the change of state of a substance.
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

d. what basis and how will you determine whether air is saturated with vapour or not?
Answer:
Whether the air is saturated with water vapour or not is determined on the basis of the extent of water vapour present in the air. If the relative humidity is 100%, the air is saturated with water vapour. In that case, we can see the formation of water droplets on the leaves of plants/grass.
If the relative humidity is less than 100%, the air is not saturated with water vapour.

Question 8.
Read the following paragraph and answer the questions:
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy.

The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses’ heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.
(1) Heat is transferred from where to where?
(2) Which principle do we learn about from this process?
(3) How will you state the principle briefly?
(4) Which property of the substance is measured using this principle?
Answer:
(1) Heat is transferred from a hot object to a cold object.
(2) This process shows the principle of heat exchange.
(3) In this process, the cold object gains heat energy and the hot object loses energy. If a system of two objects is isolated from the surroundings, heat energy lost by the hot object = heat energy gained by the cold object.
(4) This principle is used to measure the specific heat capacity of a substance.

Question 9.
Solve the following problems:
a. Equal heat is given to two objects A and B of mass 1 g. The temperature of A increases by 3 °C and B by 5°C. Which object has more specific heat? And by what factor?
Solution:
Data: m = 1 g, Δ T₁ = 3 °C, Δ T₂ = 5 °C,
Q same
Here, Q = mc₁ ΔT₁ = mc₂ ΔT₂

Thus, c₁ > c₂
The specific heat of A is more than that of B and

b. Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg, ice at 0 °C, how many grams of ammonia is to be evaporated?
(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)
Solution:
Data : m₁ = 2kg, ΔT₁=20 °C – 0 °C
= 20 °C, c₁ = 1 kcal/kg·°C, L₁ (ice) = 80 kcal/kg,
L₂ (vaporization of ammonia) = 341 cal/g = 341 kcal/kg, m₂ =?
Q₁ (heat lost by water) = m₁c₁ ΔT₁ + m₁L₁
= 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg
=40 kcal + 160 kcal = 200 kcal
Q₂ (heat absorbed by ammonia) = m₂L₂
= m₂ × 34l kcal/kg
According to the principle of heat exchange, Q₁ = Q₂
∴ 200 kcal = m₂ × 341 kcal/kg
∴ m₂ = \(\frac{200}{341}\) kg = 0.5864 kg = 586.4 g
586.4 g of ammonia are to be evaporated.

c. A thermally insulated pot has 150 g ice at temperature 0 °C. How much steam of 100 °C has to he mixed to it, so that water of temperature 50 °C will be obtained?
(Given: Latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
Solution:
Data: m₁ = 150 g, ΔT₁ = 50 °C – 0 °C
= 50 °C, c_w = 1 cal/g.°C, L₁ = 80 cal/g, L₂ = 540 cal/g,
Δ T₂ = 100°C – 50 °C = 50 °C, m₂ = ?
Q₁ (heat absorbed by ice) = m₁L₁
= 150 g × 80 cal/g = 12000 cal
Q₂ (heat absorbed by water formed on melting of ice) =m₁ c_w ΔT₁
= 150 g × 1 cal/g·°C × 50 °C = 7500 cal
Q₃ (heat given out by steam) = m₂L₂
= m₂ × 540 cal/g
Q₄ (heat given out by water formed on condensation of steam)
= m₂ c_w ΔT₂ = m₂ × 1 cal/g·°C × 50 °C
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ 12000 cal + 7500 cal = m₂ × 540 cal/g + m₂ × 50 cal/g
∴ 19500 cal = m₂ (540 + 50) cal/g
∴ m₂ = \(\frac{19500}{590}\) g
33.5 g of steam is to be mixed.

d. A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ·°C. It contains 250 g of liquid at 30 °C having specific heat of 0.4 kcal/kg·°C. If we drop a piece of ice of mass 10 g at 0 °C into the liquid, what will be the temperature of the mixture?
Solution:
Data: m₁ = 100 g, c₁ = 0.1 kcal/kg·°C,
= 0.1 cal/g·°C, T₁ = 30 °C, m₂ = 250 g,
c₂ = 0.4 kcal/kg·°C = 0.4 cal/g·°C, T₂ = 30 °C,
m₃ = 10 g, T₃ = 0 °C, L = 80 cal/g,
c (water) = 1 cal/g·°C, T = ?
Q₁ (heat lost by calorimeter) = m₁c₁ (T- T₁),
Q₂ (heat lost by liquid) = m₂c₂ (T – T₂),
Q₃ (heat absorbed by ice) = m₃ L,
Q₄ (heat absorbed by water formed on melting of ice) = m₃c (T – 0 °C)
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ m₁c₁ (T₁ – T) + m₂c₂ (T₂ – T) = m₃L + m₃c (T – 0 °C)
∴ m₁c₁T₁ – m₁c₁T + m₂c₂T₂ – m₂c₂T = m₃L + m₃c (T – 0°C)
∴ m₁c₁T₁ + m₂c₂T₂ = m₃L + (m₁c₁ + m₂c₂ + m₃c)T
∴ 100g × 0.1 cal/g°C × 30 °C + 250g × 0.4 cal/g.°C × 30 °C J
= 10 g x× 80 cal/g + (100 g × 0.1 cal/g.°C + 250 g × 0.4 cal/g.°C + 10 g × 1 cal/g.°C) T
∴ (10 + 100 + 10) T = (300 + 3000 – 800)°C
∴ 120 T = 2500 °C
∴ T = \(\frac{2500}{120}\) °C = \(\frac{125}{6}\) °C = 20.83 °C
This is the temperature of the mixture.

Project:
Take help of your teachers to make a working model of Hope’s apparatus and perform the experiment. Verify the results you obtain. [Do it your self]

Can you recall? (Text Book Page No. 62)

Question 1.
What is the difference between heat and temperature?
Answer:
Heat is a form of energy. Particles of matter (atoms, molecules, etc.) possess potential energy and kinetic energy. Total energy (potential energy + kinetic energy) of all particles of matter in a given sample is called it’s thermal energy. When two bodies at different temperatures are in thermal contact with each other, there is transfer of thermal energy from a body at higher temperature to a body at lower temperature. This energy in transfer is called heat. It is expressed in joule, calorie and erg.

Temperature is a quantitative measure of degree of hotness or coldness of a body. It is expressed in °C, °F or K (kelvin). Temperature determines the direction of energy transfer.

Question 2.
What are the different ways of heat transfer?
Answer:
Ways of heat transfer: conduction, convection and radiation.
[Note: heat ≡ heat energy. In the textbook, both the terms are used.]

Use your brain power! (Text Book Page No. 63)

Question 1.
Is the concept of latent heat applicable during transformation of gaseous phase to liquid phase and from liquid phase to solid phase?
Answer:
Yes.

Question 2.
Where does the latent heat go during these transformations?
Answer:
During these transformations, the latent heat is given out by the substance to the surroundings.

Use your brain power! (Text Book Page No. 64)

Question 1.
In the above experiment, the wire moves through the ice slab. However, the ice slab does not break. Why?
Answer:
When the thin wire with two equal weights attached to its ends is hung over the block of ice, it exerts pressure on the ice below it. Due to this, the melting point of the ice below the wire is lowered and some ice melts. The wire passes through the water so formed.

The water above the wire is no longer under pressure and, therefore, refreezes. Once again the ice below the wire melts, and the wire passes through it, and the process continues. In this way, due to alternate melting of ice and refreezing of water, the wire cuts right through the block of ice leaving the block intact.

Question 2.
Is there any relationship of latent heat with regelation?
Answer:
Yes. when the ice melts, heat is absorbed, but the temperature does not change. Also, when water refreezes, heat is given out, but the temperature does not change. This heat absorbed or given out is the latent heat.

Question 3.
You know that as we go higher than the sea level, the boiling point of water decreases. What would be the effect on the melting point of a solid?
Answer:
As we go higher than the sea level, the melting point of solids (i) that expand on melting is lowered due to a decrease in pressure (ii) that contract on melting is raised due to a decrease in pressure.

[The wire used in the experiment is made of a metal (usually copper). Metals are good conductors of heat. Hence, exchange of heat between the portion of the ice above the wire and that below the wire takes place readily.]

Can you tell? (Text Book Page No. 64)

Question 1.
We feel that some objects are cold, and some are hot. Is this feeling related in some way to our body temperature?
Answer:
Yes. If the temperature of the object is lower than our body temperature, e.g., ice, we feel the object is cold. If the temperature of the object is higher than our body temperature, e.g., hot water, we feel the object is hot.

Use your brain power! (Text Book Page No. 66)

Question 1.
How will you explain the following statements with the help of the anomalous behaviour of water?
(1) In regions with cold climate, the aquatic plants and animals can survive even when the atmospheric temperature goes below 0 °C.
(2) In cold regions in winter the pipes for water supply break and even rocks crack.
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contraeting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature or the water at the surface continues to fall to 0 °c. Finally, the water at the surface is converted Into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

(2) Sometimes water enters into crevices of the rocks. when the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts tremendous pressure on the rocks which crack and break up Into small pieces.

In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. when the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is rormed, there is an increase in the volume.

As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Fill in the blanks and rewrite the sentences:

Question 1.
The amount of water vapour in air is determined in terms of its…………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

Question 2.
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their……………
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

Question 3.
When a liquid is getting converted into solid, the latent heat is…………. (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Rewrite the following statements by selecting the correct options:

Question 1.
……….is used to study the anomalous behaviour of water.
(a) Calorimeter
(b) Joule’s apparatus
(c) Hope’s apparatus
(d) Thermos flask
Answer:
(c) Hope’s apparatus

Question 2.
When water boils and is converted into steam, then………..
(a) heat is taken in and temperature remains constant
(b) heat is taken in and temperatures rises
(c) heat is given out and temperature lowers
(d) heat is given out and temperature remains constant
Answer:
(a) heat is taken in and temperature remains constant

Question 3.
When steam condenses to form water,………..
(a) heat is absorbed and temperature increases
(b) heat is absorbed and temperature remains the same
(c) heat is given out and temperature decreases
(d) heat is given out and temperature remains the same
Answer:
(d) heat is given out and temperature remains the same

Question 4.
The temperature of ice can be decreased below 0 °C by mixing………..in it. (Practice Activity Sheet – 3)
(a) saw dust
(b) sand
(c) salt
(d) coal
Answer:
(c) salt

Question 5.
Ice/water is a substance that………..
(a) expands on melting and contracts on freezing
(b) contracts on melting and does not undergo change in volume on freezing
(c) contracts on melting and expands on freezing
(d) does not undergo any change in volume on melting or freezing
Answer:
(c) contracts on melting and expands on freezing

Question 6.
Heat absorbed when 1 g of ice melts at 0 °C to form 1 g of water at the same temperature is………..cal.
(a) 80
(b) 800
(c) 540
(d) 54
Answer:
(a) 80

Question 7.
The latent heat of vaporization of water is………..
(a) 540 cal/g
(b) 800 cal/g
(c) 80 cal/g
(d) 54 cal/g
Answer:
(a) 540 cal/g

Question 8.
The latent heat of fusion of ice is………..
(a) 540 cal/g
(b) 80 cal/g
(c) 800 cal/g
(d) 4cal/g
Answer:
(b) 80 cal/g

Question 9.
If the temperature of water is decreased from 4 °C to 10 °C, then its………..
(a) volume decreases and density increases
(b) volume increases and density decreases
(c) volume decreases and density decreases
(d) volume increases and density increases
Answer:
(b) volume increases and density decreases

Question 10.
At 4 °C, the density of water is………..
(a) 10 g/cm³
(b) 4g/cm³
(c) 4 × 10³ kg/m³
(d) 1 × 10³ kg/m³
Answer:
(d) 1 × 10³ kg/m³

Question 11.
The density of water is maximum at………..
(a) 0 °C
(b) – 4 °C
(c) 100 °C
(d) 4 °C
Answer:
(d) 4 °C

Question 12.
………..heat is needed to raise the temperature of 1 kg of water from 14.5 °C to 15.5 °C.
(a) 4180 J
(b) 103 J
(c) 1 cal
(d) 4180 cal
Answer:
(a) 4180 J

Question 13.
………..heat is needed to convert 1 g of water at 0 °C and at a pressure of one atmosphere into 1 g of steam under the same conditions.
(a) 80 cal
(b) 540 cal
(c) 89 J
(d) 540 J
Answer:
(b) 540 cal

Question 14.
Water expands on reducing its temperature below………..°C. (March 2019)
(a) 0
(b) 4
(c) 8
(d) 12
Answer:
(b) 4

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

Question 1.
Specific latent heat of fusion is expressed in g/cal.
Answer:
False. (Specific latent heat of fusion is expressed in cal/g.)

Question 2.
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on increasing.
Answer:
False. (If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C and then goes on increasing in the range 4 °C to 10 °C.)

Question 3.
At dew point relative humidity is 100%.
Answer:
True.

Question 4.
1 kcal = 4.18 joules.
Answer:
False. (1 kcal = 4180 joules.)

Question 5.
Specific heat capacity is expressed in cal/g·°C
Answer:
True.

Question 6.
Latent heat of fusion, Q = mL.
Answer:
True.

Question 7.
If the relative humidity is more than 60%, we feel that the air is humid.
Answer:
True.

Question 8.
If the relative humidity is less than 60%, we feel that the air is dry.
Answer:
True.

Question 9.
Relative humidity has no unit.
Answer:
True.

Question 10.
Absolute humidity is expressed in kg/m3.
Answer:
True.

Identify the odd one and give the reason:

Question 1.
Temperature, conduction, convection, radiation.
Answer:
Temperature. It is a physical quantity. Others are modes of transfer of heat.

Question 2.
The joule, The erg, The calorie, The newton.
Answer:
The newton. It is a unit of force. Others are units of energy (as well as work.)

Question 3.
cal/g, cal/g·°C, k cal/kg·°C, erg/g·°C.
Answer:
cal/g. It is a unit of specific latent heat. Others are units of specific heat capacity.

Match the columns:

Answer:
(1) Latent heat – Q = mL
(2) Specific heat capacity – cal/g·°C
(3) Heat absorbed or given out by a body when its temperature changes – Q = mc ΔT.

Answer the following questions in one sentence each:

Question 1.
State units of temperature.
Answer:
Units of temperature: °C, °F and K (kelvin).

Question 2.
State units of energy.
Answer:
Units of energy: the erg, the joule, the calorie.

Question 3.
State the relation between the joule and the calorie.
Answer:
1 calorie = 4.18 joules.

Question 4.
State the relation between the erg and the joule.
Answer:
1 joule = 10⁷ ergs.

Question 5.
State the relation between the erg and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10¹⁰ ergs.

Question 6.
State the relation between the joule and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10³ joules.

Question 7.
When heat energy is absorbed by an object, ΔT represents the rise in temperature. What would ΔT represent if the object loses heat energy? (Practice Activity Sheet – 4)
Answer:
If the object loses heat energy, ΔT would represent the decrease in temperature.

Answer the following questions:

Question 1.
Define latent heat of fusion.
(OR)
What is latent heat of fusion? State its units.
Answer:
When a solid is converted into liquid at constant temperature (melting point of the substance) the amount of heat absorbed by it is called the latent heat of fusion.
Heat is a form of energy. Hence, latent heat is expressed in units joule, erg, calorie or kilocalorie.

Question 2.
Define specific latent heat of fusion.
(OR)
What is specific latent heat of fusion? State its units.
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion.
It is expressed in units J/kg, erg/g, cal/g, kJ/ kg and kcal/kg.

[Note: Specific latent heat (L) = \(\frac{\text { latent heat }(Q)}{\text { mass of the substance }(m)}\)
:. SI unit of specific latent heat = SI unit of energy / SI unit of mass = J/kg]

Question 3.
Explain the term latent heat of vaporization.
Answer:
When a liquid is heated continuously, initially, its temperature increases. Later, at a certain stage, its temperature does not increase even when heat is supplied to it. At this temperature, heat absorbed by the liquid is used for breaking the bonds between its atoms or molecules, i.e., for doing work against the forces of attraction between the atoms or molecules and conversion into gaseous phase.

This heat is called the latent heat of vaporization and the constant temperature at which this change of state occurs is called the boiling point of the liquid.

Question 4.
Define boiling point of a liquid.
(OR)
What is boiling point of a liquid?
Answer:
The constant temperature at which a liquid transforms into gaseous state is called the boiling point of the liquid.
[Note: On application of pressure, the boiling point of a liquid is raised. On reducting the pressure, the boiling point is lowered.]

Question 5.
Define specific latent heat of vaporization.
OR
What is specific latent heat of vaporization?
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a liquid to convert into gaseous phase is called the specific latent heat of vaporization.

Question 6.
The specific latent heat of fusion of ice is 80 cal/g. Explain this statement.
Answer:
When 1 g of ice at a pressure of one atmosphere and at a temperature 0 °C is converted into 1 g of water, heat absorbed by the ice is 80 cal.

Question 7.
The specific latent heat of fusion of silver is 88.2 kJ/kg. Explain this statement.
Answer:
When 1 kg of silver at a pressure of one atmosphere and at a temperature of 962 °C (melting point of silver) is converted into 1 kg of silver in liquid phase, heat absorbed by the silver is 88.2 kJ.

Question 8.
The specific latent heat of vaporization of water is 540 cal/g. Explain this statement.
Answer:
When 1 g of water at a pressure of one atmosphere and at a temperature of 100 °C is converted into 1 g of steam, heat absorbed by the water is 540 cal.

Question 9.
Define regelation.
(OR)
What is regelation?
Answer:
The phenomenon in which the ice converts to liquid due to applied pressure and then re-converts to ice once the pressure is removed is called regelation.

Question 10.
The terms hot and cold are used in relative context. Explain.
Answer:

(1) Take three large bowls, P, Q and R. Fill bowl P with cold water, bowl Q with lukewarm water, and bowl R with hot water.
(2) Immerse your right hand in bowl P, and left hand in bowl R for about five seconds.
(3) Now, immerse both the hands in bowl Q at the same time.
(4) You will find that the water in bowl Q appears warm to your right hand, and cold to your left hand. Thus, the hand immersed in cold water for some time finds the lukewarm water hot while the one immersed in hot water finds the same lukewarm water cold. This experiment shows that the terms hot and cold are relative.

Question 11.
Draw a neat labelled diagram of Hope’s apparatus. Explain how this apparatus can be used to demonstrate anomalous behaviour of water. Draw a graph of temperature of water against time.
Answer:

The figure shows Hope’s apparatus. Initially, the cylindrical container in Hope’s apparatus is filled with water at about 12 °C and the flat bowl is filled with a freezing mixture of ice and salt.

The temperature of water in the upper part of the container (T₂) is recorded by thermometer T₂ and that of water in the lower part of the container (T₁) is recorded by thermometer T₁. Figure shows variation of temperature of water with time.

Initially, both the thermometers show the same temperature (say, 12 °C). In a short time, the temperature shown by the lower thermometer starts decreasing, while the temperature shown by the upper thermometer does not change very much.

This process continues till the temperature shown by the lower thermometer falls to 4 °C and remains constant thereafter. This shows that in the temperature range 12 °C to 4 °C, the density of the water in the central part of the container goes on increasing and hence the water sinks to the bottom. It means that water contracts, i.e., its volume decreases as its temperature falls from 12 °C to 4 °C.

As the temperature of the water in the central part of the container becomes less than 4 °C, the temperature shown by the upper thermometer begins to fall rapidly to 0 °C. But the temperature shown by the lower thermometer remains constant (4 °C). Later, the heading shown by the lower thermometer decreases to 0 °C.

In the temperature range 4 °C to 0 °C, the water moves upward. This shows that the density of water goes on decreasing in this range. It means that water expands, i.e. its volume increases as its temperature falls from 4 °C to 0 °C.

Thus, the volume of a given mass of water is minimum at 4 °C, i.e., the density of water is maximum at 4 °C.

In the above figure, the point of intersection of the two curves shows the temperature at which the density of water is maximum. This temperature is 4 °C.

Question 12.
A mountaineer climbing on the Everest, experienced the following facts. Explain each fact with the scientific reason : (1) He found j fishes alive below the ice (2) Time required for cooking was more as he went higher (3) He saw many times cliffs falling suddenly (4) He saw tubes carrying water broken.
Answer:
Explanation:
(1) Water expands as its temperature decreases from 4 °C to 0 °C. Water is converted into ice at 0 °C. The density of water is more than that of ice. Fishes can remain alive in the water (at 4 °C) below the ice.
(2) At high altitudes, atmospheric pressure is low and hence water boils at a temperature lower than its normal boiling point. Therefore, the time required for cooking food is more at higher altitudes.
(3) Water expands while freezing. Hence, the water present in the crevices of the rocks exerts a tremendous pressure on the rocks, while freezing. Therefore, the cliffs fall.
(4) Water expands while freezing. Hence, the water in the tube exerts a large pressure on the tube, while freezing. Therefore, the tube carrying water breaks.

Question 13.
What is humidity?
Answer:
The moisture, i.e., the presence of water vapour, in the atmosphere is called humidity.

Question 14.
When is air said to be saturated with water vapour?
Answer:
When air contains maximum possible water vapour, it is said to be saturated with water vapour at that temperature.

Question 15.
What does the amount of water vapour needed to saturate air depend on?
Answer:
The extent of water vapour needed to saturate air depends on the temperature. The greater the temperature, the greater is the amount of water vapour needed to saturate air.

Question 16.
When is air said to be unsaturated with water vapour?
Answer:
When air contains water vapour less than its capacity to hold water vapour at that temperature, it is said to be unsaturated with water’vapour.

Question 17.
What is dew point temperature?
(OR)
Define dew point temperature.
Answer:
If the temperature of unsaturated air is decreased, a temperature is reached at which the air becomes saturated with water vapour. This temperature is called the dew point temperature.

Question 18.
Name the physical quantity used to express the amount of water vapour present in air.
Answer:
Absolute humidity.

Question 19.
Define absolute humidity.
(OR)
What is absolute humidity? State its unit.
Answer:
The mass of water vapour present in a unit volume of air is called absolute humidity. Generally it is expressed in kg/m³.

Question 20.
Define relative humidity.
(OR)
What is relative humidity? Write the formula for % relative humidity.
Answer:
The ratio of the actual mass of water vapour content in the air for a given volume and temperature to that required to make the same volume of air saturated with water vapour at the same temperature is called the relative humidity.

% Relative humidity = [the actual mass of water vapour content in the air for a given volume and temperature ÷ the mass of water vapour required to make the same volume of air saturated with water vapour at the same temperature] × 100%.

Question 21.
What is the value of relative humidity at the dew point temperature?
Answer:
At the dew point temperature, relative humidity is 100%.

Question 22.
The mass of water vapour in air enclosed in a certain space is 60 g and the mass of water vapour needed to saturate the same air with water vapour under the same conditions is 100 g. What is the corresponding % relative humidity?
Answer:
Here, % relative humidity = (\(\frac{60 \mathrm{g}}{100 \mathrm{g}}\)) × 100% = 60%

Question 23.
During winter, sometimes we see a white trail at the back of a flying aeroplane in a clear sky. Explain why.
Answer:
In winter, air temperature is low. Hence, when an aeroplane flies, the vapour released by its engine condenses and forms white clouds. If the relative humidity of the air surrounding the plane is high, we see this white trail at the back of the plane for a long time before it disappears. If.the relative humidity is low, the white trail is short and disappears quickly. If the relative humidity is very low, there is no formation of the white trail.

Question 24.
State two effects of humidity present in atmosphere.
Answer:
Effects of humidity present in atmosphere: When the temperature of air falls below the dew point, dew and fog are formed.

Question 25.
Explain how dew and fog are formed.
(OR)
Write a short note on formation of dew and fog.
Answer:
At a particular temperature, a given volume of air can contain a certain maximum amount of water vapour. Normally, the temperature of air during the day is such that air is not saturated with water vapour present in it.

As the temperature falls, the capacity of air to hold water vapour becomes less. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed. If the water vapour condenses on the fine dust particles present in the atmosphere, mist or fog is formed.

Question 26.
State the units of heat.
Answer:
Units of heat: joule, erg, calorie, kilocalorie.

Question 27.
Define the kilocalorie.
Answer:
The amount of heat necessary to raise the temperature of 1 kg of water by 1 °C from 14.5 °C to 15.5 °C is called one kilocalorie.

Question 28.
Define the calorie.
Answer:
The amount of heat necessary to raise the temperature of 1 g of water by 1 °C from 14.5 °C to 15.5 °C is called one calorie.

Question 29.
State the relation between the kilocalorie and the calorie.
Answer:
1 kilocalorie = 10³ calories.

Question 30.
Study the following procedure and answer the questions below:  (Practice Activity Sheet – 2)
1. Take 3 spheres of iron, copper and lead of equal mass.
2. Put all the 3 spheres in boiling water in a beaker for some time.
3. Take 3 spheres out of the water. Put them immediately on a thick slab of wax.
4. Note the depth that each sphere goes into the wax.
(i) Which property of a substance can be studied with this procedure?
(ii) Describe that property in minimum words.
(iii) Explain the rule of heat exchange with this property.
Answer:
(i) Specific heat.
(ii) Specific heat: The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C.
(iii) According to the rule/principle of heat exchange, heat energy lost by the hot object = heat energy gained by the cold object.

In this activity, heat absorbed by the iron sphere is transmitted more in the wax, hence the sphere goes deepest into the wax, while the lead sphere absorbs less heat, resulting in less transmission of heat in the wax, hence, the sphere goes the least depth into the wax.

Question 31.
Write the symbol for specific heat capacity. State the units of specific heat capacity.
Answer:
Symbol for specific heat capacity: c.
Units of specific heat capacity: J/kg·°C,
erg/g·°C, cal/g·°C, kcal/kg·°C.
[ Notes: (1) Specific heat capacity

In SI, heat is expressed in joule (J), mass in kg and temperature in kelvin(K).
∴ SI unit of specific heat capacity = \(\frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\). (2) The specific heat capacity of a substance depends upon its constituent particles (atoms, molecules, etc.), interaction between them, structure of the substance (atomic/molecular arrangement), temperature of the substance, etc.]

Question 32.
Explain the principle of heat exchange. Ans. Suppose two objects A and B at different temperature T₁ and T₂ respectively are enclosed in a box of heat resistant material as shown in figure.

Let m₁ = mass of A, m₂ = mass of B, c₁ = specific heat capacity of A, c₂ = specific heat capacity of B and T = common temperature attained by A and B after the heat exchange between A and B. Here, no heat leaves the box or enters the box from outside. Hence, if T₁ > T₂, heat energy lost by A (Q₁) = heat energy gained by B (Q₂).
∴ m₁c₁ (T₁ – T) = m₂c₂ (T – T₂)
[Note: If m₁, c₁, T₁, T, m₂ and T₂ are known, c₂ can be determined.]

Question 33.
The specific heat capacity of silver is 0.056 kcal/kg·°C. Explain this statement.
Answer:
The amount of heat needed to raise the temperature of 1 kg of silver by 1 °C is 0.056 kcal.

Question 34.
Explain how the specific heat capacity of a solid can be determined (measured) by the method of mixture.
Answer:
A hot solid is put in water in a calorimeter. The mixture is stirred continuously and the maximum temperature of the mixture is measured with a thermometer. Heat exchange between the hot solid, water and calorimeter results in sill bodies attaining the same temperature after some time. Hence, according to the principle of heat exchange, heat lost by the solid = heat gained by the water in the calorimeter + heat gained by the calorimeter.

Now, heat lost by the solid (Q) = mass of the solid × its specific heat capacity × decrease in its temperature, heat gained by the water (Q₁) = mass of the water × its specific heat capacity × increase in its temperature and heat gained by the calorimeter (Q₂) = mass of the calorimeter × its specific heat capacity × increase in its temperature.

Heat lost by the hot object = heat gained by the calorimeter + heat gained by the water. Q = Q₂ + Q₁
Using this equation, the specific heat capacity of the solid can be determined (measured) when the other quantities are known.

Give scientific reasons:

Question 1.
Even though heat is supplied to boiling water, there is no increase in its temperature.
Answer:
Once water starts boiling, all the heat supplied to it is used in conversion of water into steam at the boiling point of water. Hence, there is no rise in its temperature.

Question 2.
Burns from steam are worse those from boiling water at the same temperature.
Answer:
1. A given quantity of steam contains more heat than the same quantity of boiling water at the same temperature.
2. When steam comes in contact with one’s body, it releases extra heat of 540 calories per gram and causes a more serious burn than that caused by boiling water.

Question 3.
In winter, the pipelines carrying water burst in cold countries.
Answer:
1. In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. When the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is formed, there is an increase in the volume.

2. As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Question 4.
If crushed ice is pressed and then the pressure is released, a lump of ice is formed.
Answer:
1. When crushed ice is pressed, its melting point is lowered and some ice melts to form water.
2. When pressure is released, the melting point becomes normal and the water freezes to form ice forming a lump.

Question 5.
In cold countries, in winter, even when the water of lakes freezes, aquatic animals and plants can survive.
Answer:
1. In cold countries, in winter, a layer of ice is formed on the surface of lakes when the atmospheric temperature falls below 0 °C. However, below this layer, there is water at 4 °C.
2. Ice, being a bad conductor of heat, does not allow transfer of heat from this water to the atmosphere. Hence, aquatic animals and plants can survive in this water.

Question 6.
Water droplets are seen on’ the outer surface of a cold drink bottle.
Answer:
1. The temperature of the outer surface of a cold drink bottle is less than that of the atmosphere.
2. Therefore, the excess of water vapour from the air condenses to form droplets on the outer surface of the cold drink bottle.

Question 7.
During cold nights, sometimes dew is formed.
Answer:
1. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. 2. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed.

Question 8.
When you enter a warm room after being outside on a frosty early morning, your spectacles ‘steam up’.
Answer:
1. On a frosty early morning, the temperature of air outside a warm room is lower than the dew point.
2. Hence, when you enter the room from outside, some water vapour in the room condenses on the glass of your spectacles, i.e., the spectacles ‘steam up’.

Question 9.
A plastic bottle, completely filled with water, when kept in a freezer, is likely to break.
Answer:
The temperature of air in the freezer (deep freeze) compartment of a refrigerator is less than 0 °C. 2. When a plastic bottle, completely filled with water, is kept in this compartment, the temperature of water falls below 4 °C and the water expands. Even when water freezes and ice is formed, there is an increase in the volume. It exerts a large pressure on the sides of the bottle and hence the bottle is likely to break.

Question 10.
The outer surface of a beaker containing ice cubes becomes wet in a short while.
Answer:
1. When ice cubes are placed in a beaker, ice starts melting. The heat required for melting is absorbed from the surrounding air and also from the beaker to some extent.
Hence, the temperature of the air and beaker falls.

2. The capacity of air to hold water vapour depends upon the temperature of the air, and this capacity decreases as the temperature decreases. At a certain low temperature, the surrounding air becomes saturated with water vapour present in it. As the temperature falls further, the air is unable to hold all the water vapour.

Hence, the extra water vapour starts condensing on the cold outer surface of the beaker in the form of minute drops. Therefore, the outer surface of the beaker containing ice cubes becomes wet in a short while.

Distinguish between the following:

Question 1.
Absolute humidity and Relative humidity.
Answer:
Absolute humidity:

1. Absolute humidity is the mass of water vapour present in a unit volume of air.
2. It is commonly expressed in kg/m³.

Relative humidity:

1. Relative humidity is the ratio of the mass of water vapour in a given volume of air at a given temperature to the mass of water vapour required to saturate the same volume of air at the same temperature.
2. It does not have unit.

Solve the following examples/Numerical problems:
[Use the data given in the Tables on pages 130 and 131.]

Problem 1.
Calculate the amount of heat required to convert 5 g of ice of 0 °C into water at 0 °C. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: Here, m = 5 g, L = 80 cal/g; Q = ?
Amount of heat required, Q = mL
= 5 g × 80 cal/g
= 400 calories.

Problem 2.
Find the amount of heat required to convert 10 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: Here, m = 10 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 10 g × 540 cal/g
= 5400 calories.

Problem 3.
Calculate the amount of heat required to convert 15 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
m = 15 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 15 g × 540 cal/g
= 8100 calories.

Problem 4.
How many calories of heat will be absorbed when 3 kg of ice at 0 °C melts?
Solution:
m = 3 kg = 3000 g; L = 80 cal/g; Q = ?
Quantity of heat absorbed, Q = mL
= 3000 g × 80 cal/g
∴ Q = 240000 calories.

Problem 5.
Calculate the amount of heat required to convert 10 g of water at 30 °C into steam at 100 °C. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
Here, m = 10 g; c = 1 cal/g·°C
T₂ – T₁ = 100 °C – 30 °C = 70 °C; L = 540 cal/g; Q = ?
Amount of heat required, Q = mc (T₂ – T₁) + mL
= 10 g × 1 cal/g·°C × 70 °C + 10 g × 540 cal/g
= 700 cal + 5400 cal
∴ Q = 6100 calories.

Problem 6.
If water of mass 80 g and temperature 45 °C is mixed with water of mass 20 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
Data : m₁ = 80 g, T₁ = 45 °C, m₂ = 20 g,
T₂ = 30 °C, T = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T1 – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₁T₁ + m₂T₂ = (m1 + m₂)T
∴ Maximum temperature of the mixture,

= (36 + 6) °C
= 42°C.

Problem 7.
When water of mass 70 g and temperature 50 °C is added to water of mass 30 g, the maximum temperature of the mixture is found to be 41 °C. Find the temperature of water of mass 30 g before hot water was added to it.
Solution:
Data : m₁ = 70 g, T₁ = 50 °C, m₂ = 30 g, T = 41 °C, T₂ = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T₁ – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₂T₂ = (m₁ + m₂) T – m₁T₁

This is the required temperature.

Problem 8.
Find the heat needed to raise the temperature of a silver container of mass 100 g by 10 °C. (c = 0.056 cal/g.°C)
Solution:
Data: m = 100 g, ΔT = 10 °C, c = 0.056 cal/g·°C
Heat needed to raise the temperature of the container = mc ΔT
= 100 g × 0.056 cal/g·°C × 10 °C .
= 56 calories.

Problem 9.
If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?
Solution:
Data: m₁ = 100 g, L₁ = 540 cal/g,
T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c (water) = 1 cal/g·°C
According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:
Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal
Decrease in the temperature of this water to 0 °C:
Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal
Melting of ice: Q₃ = mL₂
= m × 80 cal/g
Now, Q₁ + Q₂ = Q₃
∴ (54000 + 10000) cal = m × 80 cal/g
∴ m = \(\frac{64000}{80}\) = 800 g
800 g of ice will melt.

Numerical problems for practice:

Problem 1.
Calculate the amount of heat required to convert 80 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
6400 cal

Problem 2.
Find the heat required to convert 20 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
1600 cal

Problem 3.
Calculate the quantity of heat released during the conversion of 10 g of ice cold water (temperature 0 °C) into ice at the same temperature. (Specific latent heat of freezing of water = 80 cal/g)
Solution:
800 cal

Problem 4.
How many calories of heat will be absorbed when 2 kg of ice at 0 °C melts? (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
160000 cal

Problem 5.
How much heat will be required to convert 20 g of water at 100 °C into steam at 100 °C? (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
10800 cal

Problem 6.
Find the heat absorbed by 25 g of water at 100 °C to convert into steam at the same temperature. (Specific latent heat of vaporization of water = 540 cal/g.)
Solution:
13500 cal

Problem 7.
If water of mass 60 g and temperature 50 °C is mixed with water of mass 40 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
42 °C

Problem 8.
If water of mass 60 g and temperature 60 °C is mixed with water of mass 60 g and temperature 40 °C, what will be the maximum temperature of the mixture?
Solution:
50 °C

Problem 9.
Find the heat needed to raise the temperature of a piece of iron of mass 500 g by 20 °C. (c = 0.110 cal/g·°C)
Solution:
1100 cal

Problem 10.
Water of mass 200 g and temperature 30 °C is taken in a copper calorimeter of mass 50 g and temperature 30 °C. A copper sphere of mass 100 g and temperature 100 °C is released into it. What will be the maximum temperature of the mixture? [c (water) = 1 cal/g·°C, c (copper) =0.1 cal/g·°C]
Solution:
33.26 °C

Problem 11.
A copper calorimeter of mass 100 g and temperature 30 °C contains water of mass 200 g and temperature 30 °C. If a piece of ice of mass 40 g and temperature 0 °C is added to it, what will be the maximum temperature of the mixture? [c (copper) = 0.1 cal/g·°C, c (water) = 1 cal/g·°C, L = 80 cal/g]
Solution:
12.4 °C

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Question 1.
Match the pairs.

Question 2.
Draw an electron dot structure of the following molecules. (Without showing the circles)
a. Methane.
Answer:
Molecular formula: CH₄

b. Ethene.
Answer:
Molecular formula: H₂C = CH₂

c. Methanol.
Answer:
Molecular formula: H₃C – OH

d. Water.
Answer:
Molecular formula: H₂O

Question 3.
Draw all possible structural formulae of compounds from their molecular formula given below.
a. C₃H₈
b. C₄H₁₀
c. C₃H₄
Answer:
a. C₃H₈ Propane:

b. C₄H₁₀ Butane:

c. C₃H₄ Propyne:

Question 4.
Explain the following terms with example.
a. Structural isomerism.
Answer:
The phenomenon in which compounds having different structural formulae have the same molecular formula is called structural isomerism. Butane is represented by two different compounds as their structural formulae are different. The first compound is a straight chain compound and the second compound is a branched chain compound. These two different structural formulae have the same molecular formula i.e. C₄H₁₀.

b. Covalent bond.
Answer:
The chemical bond formed by sharing of two valence electrons between the two atoms is called covalent bond.
Example:
1. Hydrogen molecule formation: The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

2. Formation of oxygen molecule:
(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

c. Hetero atom in a carbon compound.
Answer:
Carbon compounds are formed by formation of bonds of carbon with other elements such as halogens, oxygen, nitrogen, sulfur. The atoms of these elements substitute one or more hydrogen atoms in the hydrocarbon chain and thereby the tetravalency of carbon is satisfied. The atom of the element which is substitute for hydrogen is referred to as a heteroatom. Sometimes hetero atoms are not alone but exist in the form of certain groups of atoms.

d. Functional group.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.

e. Alkane.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: In methane, four hydrogen atoms are bonded to carbon atom by four single covalent bonds.

f. Unsaturated hydrocarbons.
Answer:
The carbon compounds having a double bond or triple bond between two carbon atoms are called unsaturated hydrocarbons. The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
e.g. Ethene (CH₂ = CH₂), Propene (CH₃ – CH = CH₂).
The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes e.g. Ethyne (CH ≡ CH).

g. Homopolymer.
Answer:
The polymers formed by repetition of single monomer are called homopolymer. e.g. polyethylene (CH₂ – CH₂)n.

h. Monomer.
Answer:
The small unit that repeats regularly to form a polymer is called monomer.
Example: Ethylene.

i. Reduction.
Answer:
In a chemical reaction, removal of oxygen from a compound or addition of hydrogen to a compound is called a reduction.

j. Oxidant.
Answer:
An oxidant is a reactant that oxidizes or removes electrons from other reactants during a redox reaction. An oxidant may also be called an oxidizer or oxidizing agent. When the oxidant includes oxygen, it may be called an oxygenation reagent or oxygen-atom transfer (OT) agent.
Examples of oxidants include:

1. Hydrogen peroxide
2. Ozone
3. Nitric acid
4. Sulfuric acid

Question 5.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH₂ – CH₃
Answer:
The number of carbon atopic in the longest chain: 4
Parent alkane: Butane IUPAC name: n-Butane

b. CH₃ – CHOH – CH₃ (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3

Parent alkane: Propane
Functional group: -OH (ol)
Assign the number: 2
The carbon atom to which the -OH group is attached is numbered as C₂. If the carbon chain of the compound contains a -OH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘ol’. (ol stands for alcohol)
Parent suffix: Propan-2-ol
IUPAC name: Propan-2-ol

c. CH₃ – CH₂ – COOH (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane: Propane
Functional group: -COOH (-oic acid)
If the carbon chain of the compound contains a -COOH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘oic acid’.
Parent suffix: Propanoic acid
IUPAC name: Propanoic acid

d. CH₃ – CH₂ – NH₂
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -NH₂ (amine)
If the carbon chain of the compound contains a -NH₂ group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘amine’.
Parent suffix: Ethanamine
IUPAC name: Ethanamine.

e. CH₃ – CHO
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -CHO (al)
If the carbon chain of the compound contains a -CHO group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘al’.
Parent suffix: Ethanal
IUPAC name: Ethanal

f. CH₃ – CO – CH₂ – CH₃
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -CO- (one)
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the function group.
If the carbon chain of the compound contains a (-CO-) group, then change the ending of the parent name, i.e., ‘e’ of butane is replaced by ‘one’.
Parent suffix: Butan-2-one
IUPAC name: Butan-2-one

Question 6.
Identify the type of the following reaction of carbon compounds.
1. CH₃ – CH₂ – CH₂ – OH + (O) → CH₃ – CH₂ – COOH
2. CH₃ – CH₂ – CH₃ + O₂ → 3CO₂ + 4H₂O
3. CH₃ – CH = CH – CH₃ + Br₂ → CH₃ – CHBr – CHBr – CH₃
4. CH₃ – CH₃ + Cl₂ → CH₃ – CH₂ – Cl + HCl
5. CH₃ – CH₂ – CH₂ – CH₂ – OH → CH₃ – CH₂ – CH = CH₂ + H₂O
6. CH₃ – CH₂ – COOH + NaOH → CH₃ – CH₂ – COONa⁺ + H₂O
7. CH₃ – COOH + CH₃ – OH → CH₃ – COO – CH₃ + H₂O
Answer:

Question 7.
Write the structural formulae for the following IUPAC names:
a. Pent-2-one
Answer:
Pent-2-one.
(1) Pent stands for 5 carbon atoms in a chain.
Number the carbon atoms in a chain as 1, 2, 3,…..

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

b. 2-Chlorobutane
Answer:
(1) In 2-chlorobutane, butane is parent alkane stands for 4 carbon atoms and number the carbon atoms in a chain as 1, 2, 3,….

(2) Chloro (Halo) is the prefix and the number assigned for prefix (chloro) is 2. Show the chloro atom at C₂

(3) Now satisfy the valencies of each carbon atom

c. Propan-2-ol
Answer:
(1) Propan stands for 3 carbon atoms in a chain. Number the carbon atom in a chain as 1, 2, 3.

(2) ‘-ol’ stands for (-OH) hydroxyl group. The number assigned for the hydroxyl group is 2. Show the -OH group at C₂.

(3) Now satisfy the valencies of each carbon atom

d. Methanal
Answer:
(1) Meth – stands for one carbon atom and assigned the number ‘1’ to carbon in the functional group -CHO.
(2) ‘-al’ stands for functional group (-CHO) aldehyde.
(3) Now satisfy the valencies of carbon in -CHO.

e. Butanoic acid
Answer:
(1) But stands for 4 carbon atoms in a chain. Number the carbon atoms in a chain as 1, 2, 3,….

‘-oic acid’ stands for functional group -COOH. Assign the number 1 to carbon in the functional group -COOH.

Now satisfy the valencies of each carbon atom

f. 1-Bromopropane.
Answer:
(1) In 1-bromopropane, propane is parent alkane stands for 3 carbon atoms and number the carbon atoms in a chain as 1, 2, 3…….

(2) Bromo (Halo) is the prefix and the number assigned for prefix (bromo) is 1, show the bromine atom at C₁.

(3) Now satisfy the valencies of each carbon atom

g. Ethanamine
Answer:
(1) Eth stands for 2 carbon atoms in a chain and the parent alkane is ethane.
– C – C –
(2) ‘amine’ stands for (- NH₂) amino group. Show the amino (-NH₂) at any carbon atom.
– C – C – NH₂
(3) Now satisfy the valencies of each carbon atom

h. Butanone.
Answer:
(1) But stands for 4 carbon atoms in a chain and the parent alkane is butane. Number the carbon atoms in a chain 1, 2, 3,….

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

Question 8.
a. What causes the existance of very large number of carbon compound?
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(2) One, two or three covalent bonds can bond together two carbon atoms. These bonds are called single covalent bond, double covalent bond and triple covalent bond respectively. Due to the ability of carbon atoms to form multiple bonds as well as single bonds, the number of carbon compounds increases. For example, there are three compounds, namely, ethane (CH₃ – CH₃), ethene (CH₂ = CH₂) and ethyne (CH = CH) which contain two carbon atoms.

(3) Carbon being tetravalent, one carbon atom can form bonds with four other atoms (carbon or any other). This results in formation of many compounds. These compounds possess different properties as per the atoms to which carbon is bonded. For example, five different compounds are formed using one carbon atom and two monovalent elements hydrogen and chlorine: CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄. Similarly carbon atoms form covalent bonds with atoms of elements like O, N, S, halogen and P to form different types of carbon compounds in large number.

(4) Isomerism is one more characteristic of carbon compound which is responsible for large number of carbon compounds.

b. Saturated hydrocarbons are classified into three types. Write these names giving one example each.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called saturated hydrocarbons. Methane molecule contains only one carbon atom. In methane, four hydrogen atoms are bonded to carbon atom by four covalent bonds.

c. Give any four functional groups containing oxygen as the heteroatom in it. write name and structural formula of one example each.
Answer:

d. Give names of three functional groups containing three different heteroatoms. Write name and structural formula of one example each.
Answer:

e. Give names of three natural polymers. write the place of their occurance and names of monomers from which they are formed.
Answer:

1. Poly saccharide is a natural polymer. It occurs in starch/carbohydrates. It is formed from monomer glucose.
2. Protein is a natural polymer. It occurs in muscles, hair, enzymes, skin, egg. It is formed from alpha amino acids.
3. Rubber is a natural polymer. It occurs in latex of rubber tree. It is formed from monomer isoprene.

f. What is meant by vinegar and gasohol? What are their uses?
Answer:
(1) Vinegar is a 5 – 8% aqueous solution of acetic acid. It is used as preservative in pickles. It is used to cook meat. 1t is used as a salad dressing.
(2) To increase the efficiency of petrol, it is mixed with 10% anhydrous ethanol, such a fuel is called gasohol. It is used as a fuel in cars and other vehicles.

g. what is a catalyst ? write any one reaction which is brought about by use of catalyst?
Answer:
Catalyst is a substance, which changes the rate of reaction, without causing any disturbance to it. Vegetable oil (unsaturated compound) undergoes addition reaction with hydrogen in the presence of nickel catalyst to form vanaspati ghee (saturated compound).

Project:
Prepare a chart giving detailed information of carbon compounds in everyday use. Display it in the cluss and discuss.

Can you recall? (Text Book Page No. 110)

Question 1.
What are the types of compounds?
Answer:
Organic and inorganic compounds are the two important types of compounds.

Question 2.
Objects in everyday use such as foodstuff, fibres, paper, medicines, wood, fuels are made of various compounds. Which constituent elements are common in these compounds?
Answer:
The constituent elements common in these compounds are carbon (C), hydrogen (H) and oxygen (O).

Question 3.
To which group in the periodic table does the element carbon belongs? Write down the electronic configuration of carbon and deduce the valency of carbon.
Answer:
The element carbon belongs to group 14 and its electronic configuration is 2, 4. The valency of carbon is 4.

Use your brain Power! (Text Book Page No. 115)

Question 1.
The molecular formula of ethyne is C₂H₂. From this draw its structural formula and electron-dot structure.
Answer:
Ethyne: Molecular formula: C₂H₂

Question 2.
How many bonds have to be there in between the carbon atoms in ethyne so as to satisfy their tetra valency?
Answer:
To satisfy their tetravalency, three double bonds have to be there in between two carbon atoms in ethyne.

Use your brain power! (Text Book Page No. 116)

Question 1.
Draw the electron-dot structure of cyclohexane.
Answer:
Cyclohexane: Molecular formula: C₆H₁₂

Use your brain power! (Text Book Page No. 112)

Question 1.
Atomic number of chlorine is 17. What is the number of electrons in the valence shell of the chlorine?
Answer:
There are seven electrons in the valence shell of the chlorine.

Question 2.
Molecular formula of chlorine is Cl₂. Draw electron-dot and line structure of a chlorine molecule.
Answer:

Question 3.
The molecular formula of water is H₂O. Draw electron-dot and line structures for triatomic molecule. (Use dots for electron of oxygen atom and crosses for electrons of hydrogen atoms.)
Answer:

Question 4.
The molecular formula of ammonia is NH₃. Draw electron-dot and line structures for ammonia molecule.
Answer:

Question 5.
The molecular formula of carbon dioxide is CO₂. Draw the electron-dot structure (without showing circle) and line structure for CO₂.
Answer:

Question 6.
With which bond C atom in CO₂ is bonded to each of the O atoms ?
Answer:
In CO₂, carbon atom is bonded to each of the O atoms by double bond.

Question 7.
The molecular formula of sulfur is S₈ in which eight sulphur atoms are bonded to each other to form one ring. Draw electron-dot structure for S₈ without showing the circles.
Answer:

The above S₈ molecule of sulphur has crown shaped structure. One molecule of sulpur is made up of eight atoms of sulphur.

Use your brain power! (Text Book Page No. 113)

Question 1.
Hydrogen peroxide decomposes of its own by the following reaction:
2H – O – O – H → 2H – O – H + O₂
From this, what will be your inference about the strength O – O covalent bond ?
Tell from the above example whether oxygen has catenation power or not.
Answer:
In hydrogen peroxide (H₂O₂), the O – O covalent bond is not strong as oxygen has no catenation power.

Use your brain power! (Text Book Page No. 120)

Question 1.
The above table shows the homologous series of alkenes. Inspect the molecular formulae of the members of this series. Do you find any relationship, in the number of carbon atoms and the number of hydrogen atoms in the molecular formulae?
Answer:
In the above homologous series, if we observe the molecular formulae of alkenes then the number of carbon atoms are half the number of hydrogen atoms.

Question 2.
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ what will be the number of hydrogen atoms?
Answer:
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ then the number of hydrogen atoms would be 2n.

Question 3.
What would be the general formula for the molecular formulae of the members of the homologous series of alkanes? What would be the value of ‘n’ for the first member of this series?
Answer:
The general formula for the homologous series of alkane is C_nH_{2n + 2}. The value of ‘n’ for the first member of homologous series is 1.
C_nH_2n+2 = C₁H_{2 × 1} + 2 = CH₄

Question 4.
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}. Write down the individual molecular formulae of the first, second and third members by substituting the values 2, 3 and 4 respectively for ‘n’ in this formula.
Answer:
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}
n = 2 C₂H_{2 × 2 – 2} = C₂H₂ Ethyne
n = 3 C₃H_{2 × 3 – 2} = C₃H₄ Propyne
n = 4 C₄H_{2 × 4 – 2} = C₄H₆ Butyne

Question 5.
Write down structural formulae of the first four members of the various homologous series formed by making use of the functional groups.
Answer:

Question 6.
General formula of the homologous series of alkanes is C_nH_{2n + 2}. Write down the molecular formula of the 8th and 12th member using this.
Answer:
General formula of alkanes is C_nH_{2n + 2}
n = 8 C₈H_{2 × 8 + 2} = C₆H₁₈ Octane
n = 12 C₁₂H_{2 × 12 + 2} = C₁₂H₂₆ Dodecane

Use your brain power! (Text Book Page No. 121)

Question 1.
Draw three structural formulae having molecular formula C₅H₁₂. Give the names n-pentane, i-pentane and neo-pentane to the above structural formulae.
Answer:

Question 2.
Draw all possible structural formulae having molecular formula C₆H₁₄. Give names to all the isomers.
Answer:

Try This! (Text Book Page No. 124)

Question 1.
Light a bunsen burner. Open and close the air hole at the bottom of the burner by means of the movable ring around it. When do you get yellow sooty flame? When do you get blue flame?
Answer:
When the air hole at the bottom of the burner is open, sufficient oxygen is mixed gaseous fuel for complete combustion and a clean blue flame is obtained. When the air hole is partially blocked by means of the movable ring around it, the air supply is limited which results in incomplete combustion. Hence, yellow sooty flame is produced.

(Text Book Page No. 126)

Question 1.
The names of four fatty acids separated from vegetable oils are given in the table. Identify the number of carbon – carbon double bonds from their structure and molecular formula from the below fatty acids which one when reacts with iodine will make the colour of iodine disappear.
Answer:

Use your brain power! (Text Book Page No. 128)

Question 1.
Explain by writing a reaction, what will happen when pieces of sodium metal are put in n-propyl alcohol.
Answer:
n-Propyl alcohol reacts with pieces of sodium metal, sodium propoxide and hydrogen gas are obtained.

Question 2.
Explain by writing a reaction, which product will be formed on heating n-butyl alcohol with concentrated sulphuric acid.
Answer:
When n-butyl alcohol is heated with concentrated sulphuric acid, one molecule of water is removed from its molecule to form 1-butene.

Use your brain power! (Text Book Page No. 129)

Question 1.
Which one of ethanoic acid and hydrochloric acid is stronger?
Answer:
Hydrochloric acid is stronger acid.

Question 2.
Which indicator paper out of blue litmus paper and pH paper is useful to distinguish between ethanoic acid and hydrochloric acid ?
Answer:
pH paper is useful to distinguish between ethanoic acid and hydrochloric acid.

Use your brain power! (Text Book Page No. 130)

Question 1.
Explain why does the lime water turns milky in the reaction of acetic acid with sodium carbonate.
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 2.
Explain the reaction that would take place when a piece of sodium metal is dropped in ethanoic acid.
Answer:
When a piece of sodium metal is dropped in ethanoic acid, sodium acetate and hydrogen gas is formed.

Question 3.
Two test tubes contain two colourless liquids ethanol and ethanoic acid. Explain by writing reaction which chemical test you would perform to tell which substance is present in which test tube.
Answer:
Ethanol does not react with sodium bicarbonate, while ethanoic acid reacts with sodium bicarbonate to form carbon dioxide gas.

Use your brain power! (Text Book Page No. 131)

Question 1.
When fat is heated with sodium hydroxide solution, soap and glycerin are formed. Which functional groups might be present in fat and glycerin? What do you think?
Answer:
The functional group carboxylic acid (-COOH) is present in fat whereas the functioned group hydroxyl group (-OH) is present in glycerin.

Can you tell? (Text Book Page No. 131)

Question 1.
What are the chemical names of the nutrients that we get from the foodstuff, namely, cereals, pulses and meat?
Answer:
The nutrients that we get from the foodstuff, namely cereals, pulses and meat are alpha amino acids.

Question 2.
What are the chemical substances that make cloth, furniture and elastic objects?
Answer:
The chemical substances that make cloth, furniture and elastic objects are cellulose and rubber.

Use your brain power! (Text Book Page No. 133)

Question 1.
Structural formulae of some monomers are given below. Write the structural formula of the homopolymer formed from them.

Answer:

Question 2.
From the given structural formula of polyvinyl acetate, that is used in paints and glues, deduce the name and structural formula of the corresponding monomer.

Answer:

(Text Book Page No. 133)

Question 1.
Complete the following table by writing their Structural formulae and Molecular formulae.
Answer:
(Answer is given directly in bold letters.)

Fill in the gaps in the table: (Text Book Page No. 119)
(Answer is given directly in bold letters.)
a. Homologous series of alkanes.
Answer:

b. Homologous series of alcohol.
Answer:

c. Homologous series of alkenes.
Answer:

(Text Book Page No. 123)

Question 1.
Complete the table by writing the IUPAC names in the third column.
(Answer is directly given with underline.)
Answer:

Use your brain power! (Text Book Page No. 119)

Question 1.
By how many -CH₂– (methylene) units do the formulae of the first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differ? Similarly, by how many -CH₂– units do the neighbouring members ethane (C₂H₆) and propane (C₃H₈) differ from each other?
Answer:
The first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differed by one -CH₂– unit. Similarly, ethane (C₂H₆) and propane (C₃H₈) differed by -CH₂– unit.

Question 2.
How many methylene units are extra in the formula of the fourth member than the third member of the homologous series of alcohols?
Answer:
There is only one, methylene unit extra in the formula of the fourth member and the third member of the homologous series of alcohols.

Question 3.
How many methylene units are less in the formula of the second member than the third member of the homologous series of alkenes?
Answer:
There is only one methylene unit less in the formula of the second member of and the third member of the homologous series of alkenes.

Fill in the blanks and rewrite the completed statements:

Question 1.
The organic compounds having double or triple bonds in them are termed as …………
Answer:
The organic compounds having double or triple bonds in them are termed as unsaturated hydrocarbons.

Question 2.
The general formula of alkane is ……………
Answer:
The general formula of alkane is C_nH_{2n + 2}.

Question 3.
The compounds of homologous series have the same ………….. group.
Answer:
The compounds of homologous series have the same functional group.

Question 4.
A double bond is formed between carbon atoms by ………… pairs of electrons.
Answer:
A double bond is formed between carbon atoms by two pairs of electrons.

Question 5.
The compounds having different structural formulae having the same molecular formula is called ……….
Answer:
The compounds having different structural formulae having the same molecular formula is called structural isomerism.

Question 6.
The functional group of ether is …………..
Answer:
The functional group of ether is -O-.

Question 7.
The general formula of alkene is …………
Answer:
The general formula of alkene is C_nH_2n.

Question 8.
The bond between two atoms of nitrogen is a ………… bond.
Answer:
The bond between two atoms of nitrogen is a triple bond.

Question 9.
Benzene ring is made up of ………….. carbon atoms.
Answer:
Benzene ring is made up of six carbon atoms.

Question 10.
Due to …………., vegetable oil is converted into vanaspati ghee.
Answer:
Due to hydrogenation, vegetable oil is converted into vanatspati ghee.

Question 11.
………….. control the heredity at molecular level.
Answer:
Nucleic acids control the heredity at molecular level.

Question 12.
The regular repetition of a small unit is called …………..
Answer:
The regular repetition of a small unit is called polymer.

Question 13.
The structural formula of polypropylene is ……………….
Answer:
The structural formula of polypropylene is

Question 14.
The monomers of proteins are ……………..
Answer:
The monomers of proteins are alpha amino acids.

Question 15.
The monomer of cellulose is …………
Answer:
The monomer of cellulose is glucose.

Question 16.
…………. have sweet odour.
Answer:
Esters have sweet odour.

Choose the correct alternative and rewrite the statement:

Question 1.
The property of direct bonding between atoms of the same element to form a chain is called ………..
(a) catenation
(b) isomerism
(c) dehydration
(d) polymerization
Answer:
The property of direct bonding between atoms of the same element to form a chain is called catenation.

Question 2.
The molecular weight of two adjacent members in homologous series of an alkane differ by ………. units.
(a) 16
(b) 20
(c) 14
(d) 12
Answer:
The molecular weight of two adjacent members in homologous series of an alkane differ by 14 units.

Question 3.
Consecutive members of a homologous series differ by ………. group.
(a) -CH
(b) -CH2
(C) -CH3
(d) -CH4
Answer:
Consecutive members of a homologous series differ by CH₂ group.

Question 4.
……….. is used to prepare carbon black.
(a) Methane
(b) Ethene
(c) Propane
(d) Butane
Answer:
Methane is used to prepare carbon black.

Question 5.
……….. is the general formula of alkene.
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{n – 2}
Answer:
C_nH_2n is the general formula of alkene.

Question 6.
The reaction of methane with chlorine in the presence of sunlight is called ………..
(a) pyrolysis
(b) an elimination reaction
(c) a substitution reaction
(d) an addition reaction
Answer:
The reaction of methane with chlorine in the presence of sunlight is called a substitution reaction.

Question 7.
The general formula for alkynes is ………….
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{2n – 1}
Answer:
The general formula for alkynes is C_nH_{2n – 2}

Question 8.
The reaction of ………… with ethanol is a fast reaction.
(a) calcium
(b) magnesium
(c) sodium
(d) aluminum
Answer:
The reaction of sodium with ethanol is a fast reaction.

Question 9.
Ethylene has …………. bond between two carbon atoms.
(a) a single
(b) a double
(c) a triple
(d) an ionic
Answer:
Ethylene has a double bond between two carbon atoms.

Question 10.
The saturated hydrocarbons are those in which carbon atom are linked by ………….
(a) a single bond
(b) a double bond
(c) a triple bond
(d) an ionic bond
Answer:
The saturated hydrocarbons are those in which carbon atom are linked by a single bond.

Question 11.
C₇H₁₆ is ………….
(a) hexane
(b) octane
(c) methane
(d) heptane
Answer:
C₇H₁₆ is heptane.

Question 12.

(a) carboxylic acid group
(b) aldehyde group
(c) ketonic group
(d) alcohol group
Answer:
(a) carboxylic acid group

Question 13.
The possible isomers for C₅H₁₂ are ……………
(a) 2
(b) 4
(c) 1
(d) 3
Answer:
The possible isomers for C₅H₁₂ are 3.

Question 14.
………. contains alcoholic functional group.

Answer:
(d) all of these

Question 15.
Oxygen molecule has ………… bond between two oxygen atoms.
(a) a double
(b) a single
(c) a triple
(d) an ionic
Answer:
Oxygen molecule has a double bond between two oxygen atoms.

Question 16.
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be ………..
(a) colourless
(b) blue
(c) green
(d) yellow
Answer:
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be colourless.

Question 17.
Ethanoic acid has a ……… odour.
(a) rotten eggs
(b) pungent
(c) mild
(d) vinegar-like
Answer:
Ethanoic acid has a vinegar-like odour.

Question 18.
Acetic acid ………..
(a) turns red litmus blue
(b) has pungent odour
(c) is red in colour
(d) is odourless
Answer:
Acetic acid has pungent odour.

Question 19.
When acetic acid reacts with sodium metal ……….. gas is formed.
(a) oxygen
(b) hydrogen
(c) chlorine
(d) nitrogen
Answer:
When acetic acid reacts with sodium metal hydrogen gas is formed.

Question 20.
The molecular formula of acetic acid (ethanoic acid) is …………
(a) HCOOH
(b) CH₃COOH
(c) C₂H₅COOH
(d) C₃H₇COOH
Answer:
The molecular formula of acetic acid (ethanoic acid) is CH₃COOH.

Question 21.
When sodium bicarbonate solution is added to dilute acetic acid …………
(a) a gas is evolved
(b) a solid settles at the bottom
(c) the mixture becomes warm
(d) the colour of the mixture becomes yellow
Answer:
When sodium bicarbonate solution is added to dilute acetic acid a gas-is evolved.

Question 22.
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in ………..
(a) test tube A
(b) test tube B
(c) test tube C
(d) all the test tubes
Answer:
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in all the test tubes.

Question 23.
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ……….. ester is produced.
(a) ethanol
(b) ethanoic
(c) ethyl ethanoate
(d) ethyl ethanol (Practice Activity Sheet – 1)
Answer:
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ethyl ethanoate ester is produced.

Question 24.
The following structural formula belongs to which carbon compound?

(a) Camphor
(b) Benzene
(c) Starch
(d) Glucose (Practical activity sheet- 2)
Answer:
(b) Benzene

Question 25.
What type of reaction is shown below?
\(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\text { Sunlight }}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{Cl}+\mathrm{HCl}\)
(a) Addition
(b) Substitution
(c) Decomposition
(d) Reduction (Practice Activity Sheet – 3)
Answer:
(b) substitution

Question 26.
The carbon compound is used in daily life is ………..
(a) edible oil
(b) salt
(c) carbon dioxide
(d) baking soda (March 2019)
Answer:
The carbon compound is used in daily life is edible oil

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

Question 1.
Generally the melting and boiling points of carbon compounds are high.
Answer:
False. (Generally the melting and boiling points of carbon compounds are low.)

Question 2.
Till now the number of known carbon compounds is about 10 million.
Answer:
True.

Question 3.
Unsaturated hydrocarbons are less reactive than saturated hydrocarbons.
Answer:
False. (Unsatured hydrocarbons are more reactive than saturated hydrocarbons.)

Question 4.
Benzene is an aromatic compound.
Answer:
True.

Question 5.
The carbon-carbon double and triple bonds are also recognised as functional groups.
Answer:
True.

Question 6.
The general formula of alkyne is C_nH_2n.
Answer:
False. (The general formula of alkyne is C_nH_{2n – 2})

Question 7.
Naphthalene burns with a yellow flame.
Answer:
True.

Question 8.
When vegetable oil and tincture iodine react, the color of iodine does not change.
Answer:
False. (When vegetable oil and tincture iodine react, the colour of iodine changes.)

Question 9.
Saturated fats are healthy.
Answer:
False. (Saturated fats are harmful to health.)

Question 10.
Aqueous solution of ethanol is found to be neutral.
Answer:
True.

Question 11.
Denatured ethanol is used as industrial solvent.
Answer:
True.

Question 12.
Vinegar is a 12-15 % aqueous solution of acetic acid.
Answer:
False. (Vinegar is a 5-8 % aqueous solution of acetic acid.)

Question 13.
The functional group of ethanoic acid is a carboxylic group.
Answer:
True.

Question 14.
Sodium hydroxide is used in the preparation of soap from fats and oils.
Answer:
True.

Question 15.
Rubber is a manmade macromolecule.
Answer:
False. (Rubber is a natural macromolecule.)

Question 16.
Polyvinyl chloride is used in the manufacture of P.V.C. pipes and bags.
Answer:
True.

Question 17.
Polyethylene is a homopolymer.
Answer:
True.

Question 18.
The chemical bonds in carbon compounds do not produce ions.
Answer:
True.

Find the odd one out:

Question 1.
Propane, methane, ethene, pentane
Answer:
Ethene. (Others are saturated hydrocarbons.)

Question 2.
Methane, butane, benzene, sodium chloride
Answer:
Sodium chloride. (Others are organic compounds.)

Question 3.
CH₄, C₂H₆, C₃H₈, CaCO₃
Answer:
CaCO₃. (Others are organic compounds.)

Question 4.
C₂H₂, C₃H₈, C₂H₆, CH₄
Answer:
C₂H₂. (Others are saturated hydrocarbons.)

Question 5.
C₂H₄, C₄H₁₀, C₃H₈, CH₄
Answer:
C₂H₄. (Others are saturated hydrocarbons.)

Question 6.
Polyethylene, Polysaccharide, Polystyrene, Polypropylene
Answer:
Polysaccharide (Others are manmade polymers.)

Question 7.
-NH₂, -COOH,-SO₄, -Br
Answer:
-SO₄ (Others are functional groups.)

Question 8.
Methane, Ethane, Propene, Propane, Butane
Answer:
Propene (Others are members of homologous series of alkanes.)

Match the columns:

Question 1.

Answer:
(1) CH₄ – Methane
(2) Ethane – C₂H₆
(3) Alkene – CH₂ = CH₂
(4) Alkyne – C_nH_{2n – 2}.

Question 2.

Answer:
(1) Aromatic hydrocarbon – Benzene
(2) Alkane – Saturated hydrocarbon
(3) Alkyne – Propyne
(4) Alkene – C_nH_2n.

Question 3.

Answer:
(1) Cyclohexane – C₆H₁₂
(2) Methanol – CH₃OH
(3) Acetaldehyde – CH₃CHO
(4) Ethanoic acid – CH₃COOH.

Question 4.

Answer:
(1) (-OH) – Alcohol
(2) (-COOH) – Carboxylic acid
(3) (-CHO) – Aldehyde

Question 5.

Answer:
(1) Ethyne – C₂H₂
(2) Ethene – C₂H₄
(3) Ethane – C₂H₆
(4) Propyne – C₃H₄.

Question 6.

Answer:
(1) Cellulose – Wood
(2) R.N.A. – Chromosomes of plants
(3) Polyacrylonitrile – Blankets
(4) Polyvinyl chloride – P.V.C. pipes, bags.

Consider the relation between Column I and II. Fill in Column IV to match Column III.

Answer:
(1) Teflon
(2) Styrene
(3) Alpha aminoacid
(4) Nucleotide
(5) R.N.A.

Define the following:

Question 1.
Define Alkane
Answer:
Alkane: In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: Methane (CH₄), Ethane (C₂H₆)

Question 2.
Define Alkene.
Answer:
Alkene: The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
Example : Ethene (CH₂ = CH₂)

Question 3.
Define Alkyne.
Answer:
Alkyne: The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes.
Example: Ethyne C₂H₂ (CH ≡ CH).

Question 4.
Define Addition reaction.
Answer:
Addition reaction: When a carbon compound combines with another compound to form a product that contains all the atoms in both the reactants; it is called an addition reaction.

Question 5.
Define Substitution reaction.
Answer:
Substitution reaction: The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms, is called substitution reaction.

Question 6.
Define Esterification.
Answer:
Esterification: A carboxylic acid reacts with an alcohol in presence of an acid catalyst, an ester is formed. The reaction is known as esterification.

Question 7.
Define Saponification.
Answer:
Saponification: When an ester reacts with the alkali, i.e. sodium hydroxide, the corresponding alcohol and sodium salt of carboxylic acid are obtained. This reaction is called saponification reaction. It is used in the preparation of soap.
Ester + Sodium hydroxide → Sodium salt of carboxylic acid + Alcohol.

Question 8.
Define Polymerization.
Answer:
Polymerization: The reaction by which monomer molecules are converted into a polymer is called polymerization.

Name the following:

Question 1.
The higher homologue of hexane.
Answer:
Keptane.

Question 2.
The number of double bonds in benzene.
Answer:
Three.

Question 3.
The functional group in ether and halogen.
Answer:
Functional groups:
Ether: – O –
Halogen: – X (-Cl, -Br, -I).

Question 4.
Polymer of tetrafluoroethylene.
Answer:
Teflon.

Question 5.
The monomer of polysaccharide.
Answer:
Glucose.

Question 6.
The Polymer of nucleotide.
Answer:
D.N.A./R.N.A.

Question 7.
The Monomer of rubber.
Answer:
Isoprene.

Question 8.
Two oxidising compounds.
Answer:
Potassium permanganate, Potassium dichromate.

Question 9.
IUPAC name of sodium acetate.
Answer:
Sodium ethanoate.

Question 10.
The main component of natural gas.
Answer:
Methane.

Question 11.
Two isomers of butane.
Answer:
n-butane and i-butane.

Question 12.
A nomenclature system based on the structure of the compounds and it was accepted all over the world.
Answer:
International Union of Pure and Applied Chemistry (IUPAC).

Answer the following questions in one sentence each:

Question 1.
State the atomic number and electronic configuration of carbon.
Answer:
The atomic number of carbon is 6 and the electronic configuration of carbon is (2, 4).

Question 2.
State number of electrons in the outermost orbit of carbon and valency of carbon.
Answer:
Four electrons are present in the outermost orbit of carbon and the valency of carbon is 4.

Question 3.
What are hydrocarbons? Give one example.
Answer:
The compounds containing only carbon and hydrogen are called hydrocarbons. These compounds are known as organic compounds. E.g. Methane, Ethane.

Question 4.
What is the molecular formula and structural of methane?
Answer:
The molecular formula of methane is CH₄.

Question 5.
How many atoms of carbon and hydrogen are present in methane?
Answer:
The molecule of methane has one carbon atom and four hydrogen atoms.

Question 6.
State the general formula of alkane.
Answer:
The general formula of an alkane is C_nH_{2n + 2}.

Question 7.
Give two examples of alkanes.
Answer:
Methane (CH₄) and ethane (C₂H₆) are alkanes.

Question 8.
Give two examples of alkenes.
Answer:
Ethene (CH₂ = CH₂) and propene (CH₃ – CH = CH₂) are alkenes.

Question 9.
Give two examples of alkynes.
Answer:
Ethyne (HC ≡ CH) and propyne (CH₃ – C ≡ CH) are alkynes.

Question 10.
Write the name and molecular formula of a higher homologue of propane.
Answer:
Butane (C₄H₁₀) is a higher homologue of propane.

Question 11.
Write the structure and molecular formula of ethane.
Answer:
Structure of ethane:

Molecular formula of ethane: C₂H₆

Question 12.
What is meant by catenation power?
Answer:
Carbon has a unique ability to form strong covalent bonds with other carbon atoms, this result in formation of big molecules. This property of carbon is called catenation power.

Question 13.
State the structural and molecular formula of benzene.
Answer:

Question 14.
Which functional groups are present in aldehyde and ketone?
Answer:
The functional group -CHO is present in aldehyde and the functional group

is present in ketone.

Question 15.
Which functional group is present in

Answer:

Question 16.
Compare: The proportions of carbon atoms in ethanol (C₂H₅OH) and naphthalene (C₁₀H₈).
Answer:
Ethanol contains two carbon atoms while naphthalene contains 10 carbon atoms. Ethanol is a saturated hydrocarbon and naphthalene is an unsaturated hydrocarbon.

Question 17.
What are the products of combustion of methane?
Answer:
Carbon dioxide (CO₂) and water (H₂O) are the products of combustion of methane.

Question 18.
Which gas is evolved when ethanol reacts with sodium?
Answer:
Hydrogen gas (H₂) is evolved when ethanol reacts with sodium.

Question 19.
Compare: How is the transformation of ethanol into ethanoic acid on oxidation reaction?
Answer:
The transformation of ethanol into ethanoic acid is an oxidation process, in which ethanol accepts oxygen.

Question 20.
Complete the following:

Answer:

Question 21.
Which of the following hydrocarbons undergo addition reactions:
C₂H₆, C₃H₈, C₂H₄, C₃H₈?
Answer:
C₂H₄ (ethene) and C₃H₆ (propene) undergo addition reactions.

Question 22.
How many covalent bonds are there in a molecule of cyclohexane?
Answer:
A molecule of cyclohexane contains 18 covalent bonds.

Question 23.
Give the IUPAC name for CH₃COOH.
Answer:
The IUPAC name for CH₃COOH is ethanoic acid.

Question 24.
Write the IUPAC name of CH₃COONa.
Answer:
IUPAC name of CH₃COONa is sodium ethanoate.

Question 25.
What is meant by denatured alcohol?
Answer:
Ethanol is the important commercial solvent. To prevent the misuse of this solvent, it is mixed with the poisonous methanol. Such ethanol is called denatured spirit.

Question 26.
What is meant by glacial acetic acid ?
Answer:
The melting point of pure acetic acid is 17 °C. Therefore, during winter in old countries acetic acid freezes at room temperature itself and looks like ice. Therefore it is named glacial acetic acid.

Question 26.
Which useful components of hydro¬carbon are obtained by fractional distillation of crude oil?
Answer:
Various useful components of hydrocarbon such as CNG, LPG, petrol (gasoline), rockel, diesel, engine oil, lubricant, etc. are obtained by separation of crude oil using fractional distillation.

Question 28.
Which functional groups are present in ester and amine?
Answer:
Ester: -COO-
Amine: -NH₂–

Question 29.
Give two examples of natural macromolecules.
Answer:
Examples: Polysaccharide, protein and nucleic acid.

Question 30.
Write the structure of polystyrene and give its uses.
Answer:
Structure:

Polystyrene is used to make thermocoal articles.

Question 31.
Write the name and the structure of monomer of polyacrylonitrile.
Answer:
The name and structure of monomer: Acrylonitrile CH₂ = CH – CN

Question 32.
Write the name and the structure of monomer of teflon and its uses.
Answer:
The name and structure of monomer: Tetrafluro ethylene CF₂ = CF₂
Teflon is used to make nonstick cookware.

Question 33.
What is meant by copolymers?
Answer:
The polymers formed from two or more monomers are called copolymers.
Examples: Poly ethylene terephthalate.

Answer the following questions:

Question 1.
How is hydrogen molecule formed?
Answer:

The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

Question 2.
Describe the formation of oxygen molecule (O₂).
Answer:

(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

Question 3.
Describe the formation of nitrogen molecule.
Answer:

(1) The atomic number of nitrogen is 7. The electronic configuration of nitrogen is (2, 5). Nitrogen has 5 electrons in the outermost shell.
(2) It requires three more electrons to complete the L shell and attain the configuration of neon (Ne).

(3) Two nitrogen atoms come close together and share three pairs of electrons with each other, resulting in the formation of a triple bond.
(4) Thus, two nitrogen atoms are bound with a triple bond (=) to form a nitrogen molecule.

Question 4.
How is the methane molecule formed?
Answer:
(1) The electronic configuration of carbon is (2, 4). Carbon has four electrons in the outermost shell, hence it is tetravalent.
(2) The electronic configuration of hydrogen is 1, hence it is monovalent.
(3) Carbon needs four electrons to complete the L shell and attain the configuration of neon (Ne).
(4) Four atoms of hydrogen share 1 electron each with 4 electrons of carbon.
(5) A single covalent bond is formed by sharing of two electrons.

Thus, the methane molecule contains four single bonds between the carbon and hydrogen atoms.

Question 5.
State the various compounds and its formulae formed by a single atom of carbon with monovalent hydrogen and chlorine.
Answer:

Question 6.
Observe the straight chain hydrocarbons given below and answer the following questions:

(i) Which of the straight chain compounds from A and B is saturated and unsaturated straight chains?
(ii) Name these straight chains.
(iii) Write their chemical formulae and number of -CH₂ units. (Practice Activity Sheet – 2)
Answer:
(i) A is a saturated hydrocarbon, B is an unsaturated hydrocarbon.
(ii) A = Propane, B = Propene
(iii) The chemical formula of A = C₃H₈ and number of -CH₂ units are 3.
The chemical formula of B = C₃H₆ and number of -CH₂ unit is 1.

Question 7.
Draw electron-dot and line structure of an ethane molecule.
Answer:
The molecular formula of ethane is C₂H₆.

Question 8.
The molecular formula of propane is C₃H₈. From this draw its structural formula. (Practice Activity Sheet – 3)
Answer:
The structural formula of propane:

Question 9.
Draw the structure and carbon skeleton for cyclohexane.
Answer:

Question 10.
Classify into saturated and unsaturated hydrocarbons: (1) Methane (2) Ethene (3) Ethane (4) Ethyne (5) Propene (6) Propyne (7) Butane (8) Cyclohexene (9) Cyclopentane (10) Heptane.
Answer:
(i) Saturated hydrocarbons: (1) Methane (2) Ethane (3) Butane (4) Cyclopentane (5) Heptane.
(ii) Unsaturated hydrocarbons: (1) Ethene (2) Ethyne (3) Propene (4) Propyne (5) Cyclohexene.

Question 11.
Classify into alkanes, alkenes and alkynes: (1) Ethane (2) Ethene (3) Methane (4) Propene (5) Ethyne (6) Propyne (7) Butane (8) Pentane.
Answer:
Alkanes: (1) Ethane (2) Methane (3) Butane. (4) Pentane
Alkenes: (1) Ethene (2) Propene
Alkynes: (1) Ethyne (2) Propyne

Question 12.
Classify into straight chain carbon compounds, branched chain carbon compounds and ring carbon compounds:
(1) Propene (2) Butane (3) Iso-butane (4) Cyclopentane (5) Benzene (6) Isobutylene.
Answer:
Straight chain carbon compounds:

1. Propene
2. Butane.

Branched chain carbon compounds:

1. Iso-butane
2. Isobutylene.

Ring carbon compounds:

1. Cyclopentane
2. Benzene.

Question 13.
Draw chain and ring structures of organic compound having six carbon atoms in it.
Answer:
Chain structures of an organic compound having six carbon atoms:

Ring structures of an organic compound having six carbon atoms:

Question 14.
Explain the structure of benzene.
Answer:
The molecular formula of benzene is C₆H₆. It is a cyclic unsaturated hydrocarbon. Benzene ring is made of six carbon atoms. In benzene, each carbon atom is linked to two other carbon atoms, on one side by a single bond and on the other side by a double bond, i.e. three alternate single bonds and double bonds in the six membered ring structure of benzene. The compound having this characteristic unit in their structure are called aromatic compounds.

Question 15.
Draw the structures of isomers of pentane (C₅H₁₂).
Answer:

Question 16.
Recognize the carbon chain type for each of the following:

(Practice Activity Sheet – 1)
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 17.
What is meant by functional group? Give examples.
(OR)
Explain the term functional group with example.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.
Example: Methyl alcohol, acetic acid.

In methane (CH₄), when one hydrogen atom is replaced by an -OH group, methyl alcohol (CH₃OH), is formed. The -OH is known as the alcoholic functional group.
Similarly, from methane (CH₄) when one hydrogen atom is replaced by -COOH group, acetic acid (CH₃COOH) is formed. The -COOH group is known as the carboxylic acid functional group.

Question 18.
Define functional group and complete the following table:

Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the
functional groups.

Question 19.
What is meant by homologous series?
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula CnH_{2n + 2}. The members of this series are as follows:

Question 20.
State the four characteristics of homologous series.
Answer:
Characteristics of Homologous series:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit ( -CH₂– ) gets added
(b) molecular mass increases by 14 u (c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) While going in an increasing order of the length there is gradation in the physical properties i.e. the boiling and melting points.

Question 21.
Write names of first four homologous series of alcohols: (March 2019)

Answer:
First four homologous series of alcohols are

1. Methanol CH₃ – OH
2. Ethanol C₂H₅ – OH
3. Propanol C₃H₇ – OH
4. Butanol C₄H₉ – OH

Question 22.
Describe the IUPAC rules of naming organic compounds.
Answer:
IUPAC nomenclature system: International Union for Pure and Applied Chemistry (IUPAC) put forth a nomenclature system based on the structure of the compounds and it was accepted all over the world. There are three units in the IUPAC name of any carbon compound: parent, suffix and prefix. These are arranged in the name as follows:

Prefix-parent-suffix:
An IUPAC name is given to a compound on the basis of the name of its parent alkane. The name of the compound is constructed by attaching appropriate suffix and prefix to the name of the parent-alkane. The steps in the IUPAC nomenclature of straight chain compounds are as follows:

Step 1: Draw the structural formula of the straight chain compound and count the number of carbon atoms in it. The alkane with the same number of carbon atoms is the parent alkane of the concerned compound. Write the name of this alkane.

In case the carbon chain of concerned compound contains a double bond, change the ending of the parent name from ‘ane’ to ‘ene’. If the carbon chain in the concerned compound contains a triple bond, change the ending of the parent name from ‘ane’ to ‘yne’.

Step 2: If the structural formula contains a functional group, replace the last letter ‘e’ from the parent name by the condensed name of the functional group as the suffix. (Exception: The condensed name of the functional group ‘halogen’ is always attached as the prefix.)

Step 3: Number the carbon atoms in the carbon chain from one end to the other. Assign the number T to carbon in the functional group -CHO or -COOH, if present. Otherwise, the chain can be numbered in two directions. Accept that numbering which gives smaller number to the carbon carrying the functional group. In the final name, a digit (number) and a character (letter) should be separated by a small horizontal line.

Question 23.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH = CH₂
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butene
Functional group: double bond
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the double bond and the other c-atoms are numbered accordingly.
Parent suffix: But-1-ene
IUPAC name: But-1-ene

b. CH₃ – C ≡ C – H
Answer:
Number of carbon atoms in the longest chain: 3
Parent alkane: Propyne
Functional group: triple bond
Parent suffix: Propyne
IUPAC name: Propyne

c.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Prefix functional group: Chloro
Assign the number: 2

The carbon atom to which the -Cl atom is attached is numbered as C₂ and the other C atoms are numbered accordingly.
Prefix parent: 2-Chloropentane
IUPAC name: 2-Chloropentane

d. CH₃ – CH₂ – CH₂ – Br
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane : Propane Prefix functional group: Bromo
Assign the number:

The carbon atom to which the -Br atom is attached is numbered as C₁ and the other C atoms are numbered accordingly.
Prefix parent: 1-Bromopropane
IUPAC name: 1-Bromopropane

e.

Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -OH (ol)
Assign the number:

The carbon atom to which the -OH group is attached is numbered as C₂.
If the carbon chain of the compound contains a -OH group, then change the ending ‘e’ of the parent name, i.e. ,‘e’ of butane is replaced by ‘ol’ (ol for alcohol).
Parent suffix: Butan-2-ol
IUPAC name: Butan-2-ol

f.

The number of carbon atoms: 3
Parent alkane: Propane
Functional group: -NH₂ (amine)
Assign the number:

If the carbon chain of the compound contains a -NH2 group then change the ending of the parent name, i.e. ‘e’ of propane is replaced by ‘amine’.
Parent suffix: 2-Propanamine
IUPAC name: 2-Propanamine

g. HCOOH
Answer:
The number of carbon atoms: 1
Parent alkane: Methane
Functional group: -COOH (-oic cid)
If the carbon chain of the compound contains a -COOH group, then change the ending of the parent name, i.e. ‘e’ of methane is replaced by ‘-oic acid’.
Parent suffix: Methanoic acid
IUPAC name: Methanoic acid

h. CH₃ – CH₂ – CH₂ – CHO
Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane : Butane Functional group: -CHO (al)
Assign the number: 1
Assign the number ‘1’ to carbon in the functional group

If the carbon chain of the compound contains a -CHO group then change the ending of the parent name, i.e. ‘e’ of the butane is replaced by ‘al’.
Parent suffix: Butanal
IUPAC name: Butanal

i.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Functional group:

Assign the numbering:

In the longest chain, the numbering of carbon atom starts from the carbon nearest to the functional group (both the numbering equivalent).
If the carbon chain of compound contains a

group, then change the ending of the parent name i.e. ‘e’ of pentane is replaced by ‘one’.
Parent suffix: Pentan-3-one
IUPAC name: Pentan-3-one.

Question 24.
What happens when methane is burnt in air? Write the balanced chemical equation for the same.
Answer:
When methane burns in air, carbon dioxide and water are formed. The reaction is exothermic with release of large amount of heat and light.

Question 25.
What happens when ethanol is burnt in air?
Answer:
When ethanol is burnt in air, it burns with a clean blue flame, carbon dioxide and water are formed. In this reaction, release of large amount of heat and light takes place.

Question 26.
What happens when ethanol is treated with alkaline potassium permanganate? Write the balanced chemical equation for the same.
Answer:
When ethanol is treated with alkaline potassium permanganate, ethanol gets oxidised by alkaline potassium permanganate to’ form ethanoic acid.

Question 27.
What happens when vegetable oil is hydrogenated? Write the balanced chemical equation.
Answer:
When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, vanaspati ghee (saturated) compound is formed.

Question 28.
What happens when chlorine is treated with methane?
(OR)
Describe the action of chlorine on methane.
(OR)
Write a note on chlorination of methane.
Answer:
Methane reacts rapidly with chlorine in the presence of sunlight to form four products. In this reaction, chlorine atoms replace, one by one, all the hydrogen atoms in the methane.

The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms is called substitution reaction. Chlorination of methane is a substitution reaction.

Question 29.
What happens when ethanol is reacted with sodium?
Answer:
When ethanol is reacted with sodium at room temperature, sodium ethoxide is formed and hydrogen gas is liberated.

Question 30.
What happens when ethanol is heated at 170 °C with excess of conc. sulphuric acid?
Answer:
When ethanol is heated at 170 °C with excess of conc. sulphuric acid, one molecule of water is removed from its molecule to form ethene (unsaturated compound).

Question 31.
What happens when ethanoic acid is treated with sodium hydroxide? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with sodium hydroxide, neutralization takes place to form sodium acetate (sodium ethanoate) and water.

Question 32.
What happens when ethanoic acid is treated with sodium carbonate?
Answer:
When ethanoic acid is treated with sodium carbonate, sodium ethanoate, carbon dioxide and water is formed.

Question 33.
What happens when ethanoic acid is treated with sodium bicarbonate?
Answer:
When ethanoic acid is treated with sodium bicarbonate, sodium ethanoate, water and carbon dioxide is formed.

Question 34.
What happens when ethanoic acid is treated with ethanol? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with ethanol in the presence of an acid catalyst, an ester, i.e., ethyl ethanoate is formed.

Question 35.
What happens when ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst?
Ans. When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

Question 36.
State the physical properties of ethyl alcohol ethanol.
Answer:

1. Ethanol is a colourless liquid and it is soluble in water in all proportions and has pleasant odour.
2. The boiling point of ethanol is 78 °C and the freezing point is -114 °C.
3. It is combustible and burns with a blue flame.
4. An aqueous solution of ethanol is neutral to litmus paper.

Question 37.
State the properties of ethanoic acid.
Answer:

1. Ethanoic acid is a colourless liquid with boiling point 118 °C and melting point 17 °C. It has a pungent odour.
2. Its aqueous solution is acidic and turns blue litmus red.
3. A 5-8% aqueous solution of acetic acid is used as vinegar.
4. It is a weak acid.

Write short notes:

Question 1.
Catenation power.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power.

(2) Carbon shows catenation. Two or more carbon atoms can share their valence electrons and bond with each other. Thus, carbon chains can be straight or branched or closed chain ring structure forming large molecules. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(3) Hence, carbon atoms can form an unlimited number of compounds.

Question 2.
Characteristics of Carbon.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms: this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

Question 3.
Functional group.
Answer:
(1) The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of the length and nature of the carbon chain in that compound. Therefore these hetero atoms or the groups of atoms containing hetero atoms are called functional groups.

All organic compounds are derivatives of hydrocarbons. The derivatives are formed by replacing one or more H-atom/atoms of hydrocarbon by some other hetero atom or groups of atoms containing hetero atoms. After replacement, a new compound is formed which has properties different from the parent hydrocarbon.

Examples: For methane, if one hydrogen atom is replaced by an – OH group, then a compound is methyl alcohol (CH₃OH). The -OH group is known as the alcoholic functional group.
Functional group is organic compound:
1. Alcohol: – OH (hydroxy group)

4. Carboxylic acid : -COOH

Question 4.
Homologous series.
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula C_nH_{2n + 2}
The members of this series are as follows:

Characteristics:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit (-CH₂-) gets added
(b) molecular mass increases by 14 u
(c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) while going in an increasing order of the length there is gradation in the physical properties i.e., the boiling and melting points.

Question 5.
Polymerization.
Answer:
(1) The reaction by which monomer molecules are converted into a polymer is called polymerization. A macromolecule formed by regular repetition of a small unit is called polymer. The small unit that repeats regularly to form a polymer is called monomer. The important method of polymerization is to make a polymer by joining alkene type of monomers.

(2) When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

(3) The polymer polystyrene is used to make thermocoal articles. The polymer polyvinyl chloride is used to make P.V.C. pipes, doormats, etc. The polymer teflon is used to make nonstick cookware. The polymer polypropylene is used to make injection syringe, furniture, etc.

Give scientific reasons:

Question 1.
Carbon atoms are capable of forming an unlimited number of compounds.
Answer:

1. Carbon has the property of catenation. Two or more carbon atoms can share some of their valence electrons to form (single, double and triple) bonds.
2. The straight chains or branched chains or rings may have different shapes and sizes. This results in formation of many compounds. Hence, carbon atoms are capable of forming an unlimited number of compounds.

Question 2.
Ethylene is an unsaturated hydrocarbon.
Answer:
(1) Ethylene (CH₂ = CH₂) contains a double bond between carbon atoms.
(2) Thus, the valencies of the two carbon atoms are not fully satisfied by single covalent bonds. Hence, ethylene is an unsaturated hydrocarbon.

Question 3.
Naphthalene burns with a yellow flame.
Answer:
(1) Naphthalene is an unsaturated compound. In unsaturated hydrocarbon the proportion of carbon is larger than that of saturated hydrocarbon. As a result, some unburnt carbon particles are also formed during combustion of unsaturated compounds.

(2) In the flame. these unburnt hot carbon particles emit yellow light and therefore the flame appears yellow. Hence, naphthalene burns with a yellow flame.

Question 4.
The colour of iodine disappears in the reaction between vegetable oil and iodine.
Answer:
(1) Vegetable oils (unsaturated compound) contains a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.
(2) The addition reaction of vegetable oil with iodine takes place instantaneously at room temperature. The colour of iodine disappears in this reaction. This iodine test indicates the presence of a multiple bond in vegetable oil.

Question 5.
The hydrogenation of vegetable oil in the presence of nickel catalyst forms vanaspati ghee.
Answer:
(1) The molecules of vegetable oil contain long and unsaturated carbon chains. These unsaturated hydrocarbons contain a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.

(2) When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, the addition reaction takes place, vanaspati ghee (saturated compound) is formed.

Distinguish between the following:

Question 1.
Saturated hydrocarbons and Unsaturated hydrocarbons.
Answer:
Saturated hydrocarbons:

1. In saturated hydrocarbons, the carbon atoms are linked to each other only by single covalent bonds.
2. They contain only a single bond.
3. They are chemically less reactive.
4. Substitution reaction is a characteristic property of these hydrocarbons.
5. Their general formula is C_nH_{2n + 2}.

Unsaturated hydrocarbons:

1. In unsaturated hydrocarbons, the valencies of carbon atoms are not fully satisfied by single covalent bonds.
2. They contain carbon to carbon double or triple bonds.
3. They are chemically more reactive.
4. Addition reaction is a characteristic property of these hydrocarbons.
5. Their general formula is C_nH_2n or C_nH_{2n – 2}

Question 2.
Open chain hydrocarbons and closed chain hydrocarbons.
Answer:
Open chain hydrocarbons:

1. A hydrocarbon in which the chain of carbon atoms is not cyclic is called an open chain hydrocarbon.
2. All aliphatic hydrocarbons contain open chains.

Closed chain hydrocarbons:

1. A hydrocarbon in which the chain of carbon atoms is present in a cyclic form or ring form is called a closed chain hydrocarbon.
2. All aromatic hydrocarbons contain closed chains.

Question 3.
Alkane and Alkene.
Answer:
Alkane

1. Alkanes in which the carbon atoms are linked to each other only by single bonds.
2. The general formula of an alkane is C_nH_{2n + 2}
   
3. They are chemically less reactive.

Alkene:

1. Alkenes in which carbon atoms are linked to each other by double bonds.
2. The general formula of an alkene is C_nH_2n.
3. They are chemically more reactive.

Project:

Question 1.
Prepare a list of carbon compounds which occur in nature and discuss their uses in daily life.

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 8 Cell Biology and Biotechnology Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 8 Cell Biology and Biotechnology

Question 1.
Fill in the blanks and complete the statements.
a. Methods like artificial insemination and embryo transplant are mainly used for ………..
(a) animal husbandry
(b) wild life
(c) pet animals
(d) for infertile women
Answer:
(a) animal husbandry

b. ……….. is the revolutionary event in biotechnology after cloning.
(a) Human genome project
(b) DNA discovery
(c) Stem cell research
(d) All the above
Answer:
(c) Stem cell research

c. The disease related with the synthesis of insulin is …………..
(a) cancer
(b) arthritis
(c) cardiac problems
(d) diabetes
Answer:
(d) diabetes

d. Government of India has encouraged the ……….. for improving the productivity by launching NKM-16.
(a) aquaculture
(b) poultry
(c) piggery
(d) apiculture
Answer:
(a) aquaculture

Question 2.
Match the pairs.

[Note: In examination match the column question will have 2 components in Column A’ with 4 alternatives in Column B’.]
Answer:
(1) Interferon – Viral infection
(2) Factor VIII – Haemophilia
(3) Somatostatin – Dwarfness
(4) Interleukin – Cancer
[Note: Factor VIII* is an important protein factor and it should not be just factor as given in the textbook.]

Question 3.
Rewrite the following wrong statements after corrections:
a. Changes in genes of the cells are brought about in non-genetic technique.
Answer:
Non-genetic biotechnology involves use of either cell or tissue.

b. Gene from Bacillus thuringiensis is introduced into soyabean.
Answer:
Gene from Bacillus thuringiensis is introduced with gene of cotton.

Question 4.
Write short notes.
a. Biotechnology: Professional uses. (Commercial uses)
Answer:
(1) Biotechnology can be used in the following professional fields, viz. crop biotechnology, animal husbandry, human health, etc.

(2) In crop biotechnology, improvement in the yield and variety of agricultural field is done. The hybrid seeds, genetically modified crops, herbicide tolerant plants are some of the areas in which lot of biotechnological research is being done. By such research, high yielding and disease resistant varieties and varieties which can tolerate stresses such as alkalinity, weeds, cold and drought etc. are produced. BT cotton, BT Brinjal and golden rice are some GMO plants which have become popular in India.

Due to herbicide tolerant plants, the weeds are now selectively destroyed. By using biofertilizers, the use of chemical fertilizers is reduced. Use of bacteria such as Rhizobium, Azotobacter, Nostoc, Anaixiena and plants like Azolla the nitrogen fixation and phosphate solubilization abilities of the plants are improved.

(3) Animal husbandry is now using the methods of artificial insemination and embryo transfer by which the breeds of cattle are improved.

(4) To improve and to manage the human health, diagnosis ahd treatment of diseases have to be focussed. Diagnosis of diabetes, heart diseases and infectious diseases such as AIDS and dengue can be done rapidly due to biotechnology.

(5) The treatment and prevention of diseases need hormones, interferons, antibiotics and different vaccine which are now manufactured through biotechnology. Gene therapy is also used to treat hereditary disorders.

(6) Industrial products and clean technology to combat environmental pollution uses biotechnology practices.

(7) DNA fingerprinting has revolutionized the profession of forensic science.

b. Importance of medicinal plants.
Answer:

• In Ayurveda practices, the natural remedies were used. Since India had great biodiversity and traditional knowledge of herbal medicinal uses, therefore, people depended on such medicinal plants.
• In olden days, such herbs were collected by roaming in the jungles.
• Such important medicinal herbs are now cultivated with care.
• In entire world people have understood the importance of holy basil (tulsi), Adulsa, Jyesthmadh, etc.
• In some of the allopathy medicines too, the plant extracts are used.
• Medicines made from harmful chemicals have side effects and are not safe to be used unless there is medical supervision. Therefore, world-wide herbal remedies are gaining more popularity.

Question 5.
Answer the following questions in your own words.
a. Which products produced through biotechnology do you use in your daily life?
Answer:

• The simplest use of biotechnology that we practice at home is making curd and buttermilk.
• The primary type of biotechnology is used in the process of fermentation while making food stuffs, like bread, idli-dosa, dhokla, etc.
• Nowadays, different types of cheese, paneer, yoghurt, energy drinks, etc. are produced with the help of biotechnology. We are consuming these in our daily life.
• Seedless grapes, papaya, and watermelons are available in the market these days.
• Violet cabbage, yellow capsicum and exotic vegetables used for salad are also biotechnology products.
• The vaccines, antibiotics and the injections of human insulin are in regular use in many house-holds.

b. Which precautions will you take during spraying of pesticides?
Answer:

• Pesticides are toxic chemicals. By using them indiscriminately, they contaminate the water, soil and also crops.
• The D.D.T., chloropyriphos and malathion are very dangerous. They spread through the food chain causing biomagnification.
• Therefore, we shall not use such insecticides and pesticides. We shall use organic pesticides. Excessive use will be avoided.
• At the time of spraying, nose, eyes and skin will be covered and protected.
• Care will be taken not to allow children or domestic animals to come in, contact with a pesticide.

c. Why some of the organs in human body are most valuable?
Answer:

• The body can be in best health,if all the vital organs of the body are also in the best condition.
• Brain, kidney, heart, liver, etc. are some such vital organs which are most essential for proper metabolism and functioning of the body. The sense organs of the body are also of utmost importance, especially eyes.
• One cannot survive if any of these vital organs are not functioning properly. Some of the organs like brain will never regenerate too.
• Some of the organs can be brought back to functionality by performing surgeries. However, any problem with these vital organs make life miserable, therefore, they are said to be valuable.

d. Explain the importance of fruit processing in human life?
Answer:
(1) Fruits are perishable food stuff. They are spoilt soon if not consumed immediately. Hence for storage and usage for a long term, their preservation is absolutely essential.
(2) For year-long use of the fruits they are dried, salted, packed in air tight containers, used for preparing jams and jellies or condensed into pulps or syrups. Beverages, pickles, sauce, and various other products made from the fruits are largely used by us.
(3) The preserved products also fetch financial benefits.
(4) In national and international markets, Indian fruits like mangoes are in great demand. We can get foreign currency through exports of fruits and fruit products. The local horticulturists get good benefit from their orchards.
(5) Processed fruit products also give vitamins and minerals that help in maintaining good health. Thus fruit processing is important for human life.

e. Explain the meaning of vaccination.
Answer:

• Vaccination is the administering of vaccine. Vaccine is the ‘antigen’, given to a person or even to animals for acquiring immunity against particular pathogens or diseases.
• In olden days, vaccipes were prepared with the help of completely or partially killed pathogens. But this method causes some inconvenience. Some persons were allergic to such raw vaccines or they contracted the same disease through such vaccines.
• Hence in recent times the vaccines are produced by using biotechnology. These vaccines are artificial which are synthesised in the laboratories.
• The antigen is produced with the help of gene of the pathogen. Such vaccine becomes safe for administering.
• These antigenic proteins are injected to people to make their immune systems strong. This process of vaccination is absolutely safe. The vaccines are more thermostable and active for a long period of time.

Question 6.
Complete the following chart.

Answer:

Question 7.
Write the correct answer in blank boxes.

Answer:

Question 8.
Identify and complete the following correlations:
a. Insulin : Diabetes : : Interleukin : …………….
Answer:
Insulin : Diabetes : : Interleukin : Cancer

b. Interferon : ………. : : Erythropoietin : Anaemia.
Answer:
Interferon : Viral infection : : Erythropoietin : Anaemia.

c. ……….. : Dwarfness : : Factor VIII : Haemophilia.
Answer:
Somatostatin : Dwarfness : : Factor VIII : Haemophilia.

d. White revolution : Dairy : : Blue revolution : ………
Answer:
White revolution : Dairy : : Blue revolution : Fishery

Question 9.
Write a comparative note on usefulness and harmfulness of biotechnology.
(OR)
“Biotechnology is not only beneficial but it has some harmful effects too”. Express your opinion about this statement.
Answer:
(1) Biotechnology has proved to be useful in the field of agriculture, medicine, clean technology and industrial products.
(2) Due to various biotechnological experiments, the food production is increased substantially. The milk and milk products are now freely available. People no longer die of hunger due to abundant food supply.
(3) The sophisticated vaccines have stopped the spread of epidemics.
(4) The diseases like diabetes can be controlled due to human insulin injections that can be manufactured by biotechnology.
(5) The problems of pollution control, solid waste management and fuels are partially tackled by biotechnological alternatives.
(6) Though all such positive aspects are there, the biotechnology also poses some problems. The genetic changes are breaking the principles of nature. By inserting human genes in bacteria or virus, the products that are needed only for humans are produced.
(7) Human cloning is also a debatable issue. It will cause social and ethical problems. The new generations formed by cloning will have mothers but no fathers. If man tries to manipulate the genomes of other living organisms, it will cause disturbances in the natural balance. The long ternT effects of all such genetic manipulations can be disastrous. Thus, according to some views, biotechnology can be dangerous too.

Projects: (Do it your self)

Project 1.
Visit the organic manuring projects nearby your place and collect more information.

Project 2.
What will you do to increase public awareness about organ donation in your area?

Project 3.
Collect information about ‘green corridor’. Make a news-collection about it.

Can you recall? (Text Book Page No. 88)

Question 1.
What is cell?
Answer:
The structural and functional unit of the body is called a cell.

Question 2.
What is tissue? What are the functions of tissue?
Answer:
Tissue is a group of cells that performs a similar and definite function. E.g. The muscular tissues in the body perform contraction and extensions thereby helping in locomotion. The conducting tissues of the plants like xylem and phloem transport the water and food respectively.

Question 3.
Which technique in relation to tissues have you studied in earlier classes?
Answer:
The technique of tissue culture and genetic engineering has been studied last year. Tissue culture is ‘Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’. Genetic engineering and its use has also been studied under, ‘Introduction to biotechnology’.

Question 4.
Which are the various processes in tissue culture?
Answer:
Various step-wise processes are done while performing the-tissue culture. These processes are primary treatment, reproduction/cell division/multiplication, shooting or rooting, primary hardening, secondary hardening, etc.

Observe: (Text Book Page No. 88)

Question 1.
Assign names in the figure given below. Explain the various stages those are kept blank:

Answer:

Tissue Culture: Tissue culture is the technique in which ‘ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’ is done. While performing experiments of tissue culture, a liquid, solid or gel-like ” medium prepared from agar, is used. Such medium supplies nutrients and energy necessary for tissue culture technique. Different processes are to be done while performing tissue culture, viz. primary treatment, reproduction or multiplication, shooting and rooting, primary hardening, secondary hardening, etc. From the source plant, required tissues are taken out and all the processes are carried in an aseptic medium in laboratory.

(Use your brain power. (Text Book Page No. 89)

Question 1.
Just like the grafting in plants, is the organ transplantation possible in humans?
Answer:
The grafting as done in case of plants, cannot be done in human beings. But the transplantation of certain organs can be done. Liver, kidney, heart, eyes, etc. can be transplanted. But for these transplantations the donor and the recipient should match with each other in respect of their bloodr groups, age, disease condition, etc. In future, the stem cell research can bring about certain changes in the field of transplantations.

(Text Book Page No. 94)

Question 1.
What will happen if the transgenic potatoes are cooked before consumption?
Answer:
Some types of transgenic potatotes that contain edible vaccine against Hepatitis can be cooked. The cooking does not destroy the antigen incorporated into these transgenic potatoes. But according to some scientists, transgenic potatoes with enterotoxin vaccine, if cooked shows denaturation of vaccine.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
The property of stem cells is called ………….
(a) diversity
(b) equality
(c) differentiation
(d) pluripotency
Answer:
(d) pluripotency

Question 2.
Cell ……….. starts from 14^th day of conception.
(a) development
(b) specialization
(c) growth
(d) differentiation
Answer:
(d) differentiation

Question 3.
Availability of ………… is an important requirement in organ transplantation.
(a) doctor
(b) clinic
(c) donor
(d) ambulance
Answer:
(c) donor

Question 4.
The toxin which is lethal for ……….. was produced in leaves and bolls of BT cotton.
(a) bollworm
(b) locust
(c) birds
(d) frogs
Answer:
(a) bollworm

Question 5.
Transgenic raw potatoes generate the immunity against ………… disease.
(a) plague
(b) cholera
(c) leprosy
(d) TB
Answer:
(b) cholera

Rewrite the following wrong statements after corrections:

Question 1.
High-class varieties of crops have been developed through the technique of transplantation.
Answer:
High-class varieties of crops have been developed through the technique of tissue-culture.

Question 2.
Earlier, insulin was being collected from, the pancreas of pigs.
Answer:
Earlier, insulin was being collected from the- pancreas of horses.

Question 3.
Malaria arises due to genetic changes in hepatocytes.
Answer:
Phenylketonuria (PKT) arises due to genetic changes in hepatocytes.

Question 4.
The E.coli bacteria are useful for cleaning the hydrocarbon and oil pollutants from soil and water.
Answer:
The Pseudomonas bacteria are useful for cleaning the hydrocarbon and oil pollutants from soil and water.

Question 5.
Various essential elements like N, P, K are removed and hence become unavailable to the crops due to earthworms and fungi.
Answer:
Various essential elements like N, P, K become available to crops due to earthworms and fungi.

Question 6.
We do not have any tradition that cures the diseases with the help of natural resources.
Answer:
We have a great tradition of ayurveda that cures the diseases with the help of natural resources.

Match the pairs:

Question 1.

Answer:
(1) Dr. Anand Mohan Chakravarti – Cleaning the oil spill
(2) Dr. M.S. Swaminathan – Green revolution in India

Question 2.

Answer:
(1) Pseudomonas – Hydrocarbons
(2) Pteris vitata – Arsenic

Find the odd man out:

Question 1.
Green revolution, Industrial revolution, White revolution, Blue revolution
Answer:
Industrial revolution. (All others are concerned with food.)

Question 2.
DDT, malathion, chloropyriphos, Humus
Answer:
Humus. (All others are insecticides.)

Question 3.
Sodium, Aluminium, Potassium, Phosphorus
Answer:
Aluminium. (All others are essential elements for plant growth.)

Question 4.
Diabetes, Anaemia, Leukaemia, Thalassemia
Answer:
Diabetes. (All other diseases involve reduction in the number of blood cells.)

Question 5.
Drying, Salting, Cooking, Soaking with sugar
Answer:
Cooking. (All others are food preservative methods.)

Identify and complete the following correlations:

Question 1.
White revolution : Increase in dairy production : : Green revolution : ………. (March 2019)
Answer:
White revolution : Increase in dairy production : : Green revolution : Increase in agricultural production or crop yield

Question 2.
Nostoc, Anabaena : Biofertilizers : : Alfalfa : ………..
Answer:
Nostoc, Anabaena : Biofertilizers : : Alfalfa : Phytoremediation.

Give definition/Give meanings:

Question 1.
Stem cell or what are stem cells?
Answer:
The special cells having pluripotency and ability to divide and differentiate into new cells are called stem cells. They are present in multicellular living beings.

Question 2.
Biotechnology.
Answer:
Technology that brings about artificial genetic changes and hybridization in organisms for human welfare is called biotechnology.

Question 3.
Genetically modified crops.
Answer:
Crops having desired characters are developed by integrating foreign gene with their genome, such crops have modified genome and are known as genetically modified crops.

Question 4.
Golden rice.
Answer:
Biotechnologically developed variety of rice in which gene synthesizing the vitamin A (Beta carotene) has been incorporated and which contains 23 times more amount of beta carotene than that of the normal variety is called golden rice. It was developed in 2005.

Question 5.
Vaccine.
Answer:
The ‘antigen’ containing material given to a person or animal to acquire either permanent or temporary immunity against a specific pathogen or disease is called a vaccine.

Question 6.
Cloning.
Answer:
Production of replica of any cell or organ or entire organism through biotechnological process is called cloning.

Question 7.
DNA fingerprint.
Answer:
The nucleotide sequence present on the DNA of each person is unique just like the fingerprint, thus for establishing the identity of any person DNA can be analysed, this technique is known as DNA fingerprinting.

Question 8.
Green revolution.
Answer:
All the methods applied for harvesting maximum yield from minimum land are collectively referred to as green revolution.

Question 9.
White revolution.
Answer:
Achieving the self-sufficiency in dairy business, by performing various experiments for quality control, bringing about newer dairy products and their preservation and thus raising economic standards is called white revolution.

Question 10.
Blue revolution.
Answer:
The aquaculture practices to increase the yield of edible aquatic organisms is called blue revolution.

Name the following:

Question 1.
Research institutes involved with cell science.
Answer:

• National Centre of Cell Science, Pune
• Instem, Bengaluru.

Question 2.
Sources of stem cells.
Answer:

• Umbilical cord
• Embryonic cells
• Redbone marrow
• Adipose connective tissue and blood of adult human being.

Question 3.
Types of Stem cells.
Answer:

• Embryonic stem cells
• Adult stem cells.

Question 4.
Organs that can be donated.
Answer:
Eyes, heart, pancreas, liver, kidneys, skin, J bones, lungs.

Question 5.
Organisms used as biofertilizers.
Answer:
Rhizobium, Azotobacter, Nostoc, Anabaena, Azolla.

Question 6.
Two main methods used in animal husbandry.
Answer:

1. Artificial insemination
2. Embryo transfer.

Question 7.
Two important aspects of human health management.
Answer:

1. Diagnosis
2. Treatment of diseases.

Question 8.
Place where DNA fingerprinting research is done in India.
Answer:
Centre of DNA fingerprinting and Diagnostics, Hyderabad.

Question 9.
One benefit of biotechnology to the agriculture.
Answer:
Expenses on the pesticides are reduced.

scientific reasons:

Question 1.
Nowadays, safer vaccines are being produced.
Answer:

• Before the advent of biotechnology, the vaccines were made from inactive or dead pathogens of that disease.
• But now the vaccine is made artificially using biotechnological processes.
• Such vaccines produced some disease symptoms in some cases.
• The antigen of the disease is researched upon and its genetic code is found out.
• A similar antigen is made in the laboratories which is used as a vaccine.
• Such vaccines are more thermostable and remain active for longer duration. Therefore, the vaccines are now safer.

Question 2.
Awareness about organ donation after death is increasing.
Answer:

• Due to accidents or illness, some of the vital organs may get damaged and may not work to fullest capacity.
• In such cases, if organ transplantation is done, it will be very helpful for that needy patient.
• The dead person’s organs can be used for organ transplantation and a life can be saved.
• Many government and social organizations are spreading awareness about such donations. Therefore, gradually the awareness about organ transplantation is increasing.

Answer the following questions:

Question 1.
Write two uses of biotechnology related to human health. (Board’s Model Activity Sheet)
Answer:

1. Biotechnology is used to manufacture vaccines for controlling diseases.
2. Different hormones such as insulin, somatotropin and somatostatin can be prepared in laboratories by using new biotechnological processes. The clotting factors are also manufactured through such techniques.

Question 2.
Answer the following questions:
(a) What is biotechnology?
(b) Explain any two commercial applications of it. (March 2019)
Answer:
(a) Biotechnology: Technology that brings about artificial genetic changes and hybridization in organisms for human welfare is called biotechnology.

(b)

• The treatment and prevention of diseases need hormones, interferons, antibiotics and different vaccine which are now manufactured through biotechnology. Gene therapy is also used to treat hereditary disorders.
• Industrial products and clean technology to combat environmental pollution uses biotechnology practices.
• DNA fingerprinting has revolutionized the profession of forensic science.

Question 3.
What is mainly included under biotechnology?
Answer:
Biotechnology includes the following main areas:

• Abilities of microbes are used in producing yoghurt from milk and making alcohol from molasses.
• Production of antibiotics and vaccines, etc. is carried out by with the help of specific cells using their productivity.
• Bio-molecules like DNA and proteins are used for human welfare.
• By performing gene manipulation, plants, animals and products of desired quality are produced. Genetically modified bacteria are used to produce human hormones such as Human Growth Hormone and insulin.
• Tissue culture is a non-genetic technique which is used for production of new cells or tissues. Hybrid seeds are also produced in a similar way.

Question 4.
What are edible vaccines?
Answer:

• Edible vaccines are those which are given as a food by incorporating them into the food-stuff.
• Such edible vaccines are produced through biotechnology.
• Transgenic potatoes are produced with the help of biotechnology which contain vaccine that act against bacteria like Vibrio cholera, Escherichiatoli.
• If raw potatoes are consumed, then the immunity is generated in the body of a person. However, eating only raw potatoes generates the immunity against cholera and the disease caused due to E. coli.

Question 5.
What is DNA fingerprinting? Explain it in brief. Where is this technique used? Give any two examples. (Board’s Model Activity Sheet)
Answer:

• As the fingerprints are unique for every individual, similarly the nucleotide sequence in the DNA molecule is also unique.
• By knowing this sequence, one can find out the identity of any person. Such technique to establish the identity of a person by taking into consideration the nucleotide sequence is called DNA fingerprinting.
• Its main use is in forensic sciences to confirm the identity of the criminal.
• Similarly, identity of parents in case of disputed parentage for any child can be understood by taking DNA fingerprints of both the parents and a child.

Write short notes on:

Question 1.
Uses of stem cells.
Answer:
Stem cells are used for following purposes:

• In regenerative therapy stem cells are used.
• In case of diseased conditions like diabetes, myocardial infarction, Alzheimer’s disease, Parkinson’s disease, etc., stem cells can be used to replace the damaged or functionless cells.
• In conditions such as anaemia, thalassaemia, leukaemia, etc. there is always the need of newer blood cells. Here, stem cells can be used to restore the number of blood cells.
• In techniques of organ transplantation stem cells can be used and they can help in the transplantation of new organs such as kidney and liver The defective organs can be replaced by those that are produced with the help of stem cells and transplanted.

Question 2.
Cloning.
Answer:

• Cloning is the modern technique in which there is production of replica of any cell or organ or entire organism is done.
• There are two types of cloning, viz. (i) Reproductive cloning and (ii) Therapeutic cloning.
• Reproductive cloning: In reproductive cloning, a clone is produced by fusion of a nucleus of diploid somatic cell with the enucleated ovum of anybody. In the process, the sperm or male gamete is not needed.
• Therapeutic cloning: This technique is largely used for treatment purpose. Stem cells are derived from the cell formed in laboratory by the union of somatic cell nucleus with the enucleated egg cell.
• This technique is used for therapy of various diseases.
• Gene cloning can also be done to form millions of copies of same gene. Such genes are used for gene therapy and other purposes.
• Due to cloning technique, the inheritance of hereditary diseases can be controlled, continuation of generations can be achieved and certain characteristic genes can be enhanced.
• However, for human cloning, there is world-wide opposition due to ethical reasons.

Question 3.
Dolly.
Answer:

• Dolly was the first mammalian cloned sheep.
• Dolly was born on 5th July 1996 in Scotland by the process of cloning.
• The Finn Dorset sheep was chosen and her diploid nucleus from the udder cell was introduced into the ovum whose haploid nucleus was removed. This enucleated ovum was of Scottish sheep.
• The egg was then introduced into uterus of another Scottish sheep and it grew into Dolly.
• Dolly resembled exactly like Finn Dorset sheep whose diploid nucleus was used. None of the characters of Scottish sheep were seen in Dolly.
• In this way, Dolly had three mothers but no father.
• Dolly gave birth to many young ones. She died on 14th February 2003 due to cancer of the lungs.

Question 4.
Green revolution.
Answer:

• In agriculture, different methods used to harvest maximum yield from minimum land, these methods are collectively called green revolution.
• Dr. M.S. Swaminathan is called father of Green Revolution in India while Dr. Norman Borlaug has done the similar efforts in the U.S.
• Before the Green Revolution in India, there was always the dearth of the food grains. The overflowing Indian population was badly affected due to poor quality and quantity of food.
• But due to the Green Revolution in India, attention was focussed on the agricultural research.
• Improvised dwarf varieties of wheat and rice, proper use of fertilizers and pesticides and water management were the proper methods that increased production of food grains.
• This created abundance of the grains for Indian population.

Question 5.
White revolution.
Answer:

• Few years back, there was scarcity of milk in various parts of India. At some places, milk and milk products were abundant but they did not reach all the consumers.
• Dr. Verghese Kurien ^ho was then the founder director of Anand Milk Union Limited (AMUL) started thecooperative movement in the direction to produce “operation flood”, i.e. abundance of milk everywhere.
• The use of biotechnology was also done to increase the milk production.
• Dr. Kurien’s efforts have reached all-time high status as India is now self-sufficient in dairy business.
• This is popularly known as White Revolution. Different experiments were performed for quality control, newer dairy products were thought off and preservation methods were improved.
• This created White Revolution. AMUL from Anand has now reached international standards.

Question 6.
Blue Revolution.
Answer:

• Utilization of aquaculture practices for obtaining edible and commercial aquatic organisms is called blue revolution.
• In East Asian countries where water bodies and fish population is abundant, the aquaculture was started.
• On similar lines, in India, the aquaculture of different fresh water and marine organisms is being done with the help of fishery scientists.
• Government of India has vowed to increase the aquaculture production by encouraging the people for aquaculture by launching the program ‘Nil- Kranti Mission-2016’ (NKM-16).
• Pisciculutre is culturing of fish, mariculture is culture of marine organisms such as prawns/shrimps and lobsters. Sea weeds, oysters, clams are also cultured.
• For carrying out aquaculture, 50% to 100% subsidies are offered by the Government.
• Fresh water fishes like rohu, catla and other edible varieties like shrimp and lobsters are being cultured on a large scale which can bring about Blue Revolution.

Complete the paragraph by choosing the appropriate words given in the bracket:

Question 1.
(degenerated, red bone marrow, adipose connective tissue, blastocyst, umbilical cord, Differentiation)
………… of stem cells form can form various tissues, in the body. Stem cells are present in the ………….. by which the foetus is joined to the uterus of the mother. Stem cells are also present in the ……….. stage of embryonic development. Stem cells are present in ……….. and ………… of adult human beings. It has become possible to produce different types of tissues and the ……… part of any organ with the help of these stem cells.
Answer:
Differentiation of stem cells form can form various tissues in the body. Stem cells are present in the umbilical cord by which the foetus is joined to the uterus of the mother. Stem cells are also present in the blastocyst stage of embryonic development. Stem cells are present in red bone marrow and adipose connective tissue of adult human beings. It has become possible to produce different types of tissues and the degenerated part of any organ with the help of these stem cells.

Paragraph-based questions :

1. Green corridor refers to a special road route that enables harvested organs meant for transplants to reach the destined hospital. A 45-year-old woman, a victim of a railway accident, was declared brain dead, her husband and children agreed to donate her kidneys, liver and heart. One of her kidneys was transplanted to a patient in MGM Hospital and the second kidney helped a patient in Jaslok hospital. Her liver helped the transplant of a patient in Wockhardt Hospital. And her heart was sent to Fortis to the patient on a super urgent priority list, transported via a green corridor covering 18km in less than 16 minutes. This was possible due to Green corridor.
Questions and Answers :

Question 1.
What is Green corridor?
Answer:
Green corridor is a special road route that enables harvested organs meant for transplants to reach the destined hospital

Question 2.
Which organs of brain-dead lady were transplanted?
Answer:
Two kidneys, liver and heart of the brain- dead lady were transplanted.

Question 3.
How many lives were saved from organs of one lady?
Answer:
Four patients lives were saved due to organ donation of one lady.

Question 4.
How was distance of 18km covered in 16 minutes? Why?
Answer:
The distance was covered because the concept of Green corridor was applied. The heart was sent from one hospital to another, where the recipient was kept ready. The quick transportation is necessary to keep heart in living condition.

Question 5.
Who takes the decision to donate the organs?
Answer:
The close relatives of deceased person take the decision to donate the organs.

2. Read the following extract and answer the questions that follow: (March 2019)
A liberal view behind the concept of organ and body donation is that after death our body should be useful to other needful persons so that their miserable life would become comfortable. Awareness about these concepts is. increasing in our country and people are voluntarily donating their bodies.

Life of many people can be saved by organ and body donation. Blinds can regain their vision. Life of many people can be rendered comfortable by donation of organs like liver, kidneys, heart, heart valves, skin, etc. Similarly, body can be made available for research in medical studies. Many government and social organizations are working towards increasing the awareness about body donation.
Questions and Answers :

Question 1.
What is the liberal view behind the organ and body donation?
Answer:
By body donation, research in medical studies is possible. The needy persons can get vital organs which can save their lives.

Question 2.
Name any four organs that can be donated.
Answer:
Liver, Kidneys, heart, eyes, skin, etc. can be donated.

Complete the following table:

Question 1.

Answer:

Diagram/chart based questions:

Question 1.

(A) Which process is shown in the above figure? *
Answer:
The figure shows process to make transgenic

(B) Describe in brief the steps I, II, III and IV.
Answer:

Question 2.
Draw well labelled diagram of Stem cell therapy.
Answer:

Question 3.
Label the following diagram :
(i) Stem cells and organ transplantation,
Answer:

(ii) Organs that can be donated:
Answer:

Question 4.
(i) Which therapy is shown in the Fig. 8.5?

(ii) Which will be possible benefits of this therapy in organ transplantation ?
Answer:
(i) The figure 8.5 shows the ‘regenerative therapy’ using stem cells. Also called stem cell therapy.
(ii) With the help of above therapy organs like liver, kidney from stem cells can be redeveloped to replace the failed ones.

Activity based questions:

Question 1.
Bring a packet of ‘Balghuti’ from ayurveda shop. Learn the information about each component in it. Collect information about various other medicines and prepare the chart as shown below. (Try this: Textbook page no. 99)
Answer:

Question 2.
Give five examples of each of the fruiting and flowering plants developed through tissue culture and mention their benefits. (Make a list and discuss: Textbook page no. 93)
Answer:
I. Fruiting trees: Banana, Chikoo (Sapota), Tomato, Fig, Pineapple.

II. Flowering trees: Orchids, Roses, Chrysanthemum, Gerbera, Begonia, Carnation, Lili. Benefits of such plants may be varied. Mostly fruits developed are made seedless and tastier.

III. Benefits of plants produced through technique of tissue culture:

• Techniques of tissue culture can produce more copies of same plant with better characters. ’ The plant grower likes to have bigger and more fruits from fruit trees. On the flowering trees, colourful flowers with good fragrance are favoured.
• Plants which do not depend on particular climate and local seasonal changes are produced by tissue culture methods. This helps to rise the yield in an area which otherwise may not produce a specific crop.
• For tissue culture, saplings and seedlings are made available throughout the year through laboratory. The limitations of getting natural seeds are not there thus planting can be done throughout the year.
• Tissue culture techniques create the plants of uniform size, shape and yield. Since they are exactly alike, it becomes beneficial.
• In lesser time period, the crops reach maturity.
• The crops are pest and disease resistant.
• Tissue culture techniques are cost effective and easy to carry out.

Question 3.
Which new species of the rice have been developed in India? (Collect Information: Textbook page no. 97)
Answer:

1. Species in 2015-16: High zinc species (DRR Dhan 45), Pusa 1592, Punjab basmati 3, Pusa 1609, Telangana Sona.
2. Species in 2014: CR Dhan 205, CR Dhan 306, CRR, 451.

Question 4.
Discuss about stem cells and organ transplantation in the class with the help of figures given on textbook page no. 90. (Observe: Textbook page no. 90)
Answer:


Organ transplantation:
Various organs in the human body either become less efficient or completely functionless due to various reasons like aging, accidents, infections, disorders, etc. Life of such person becomes difficult or even fatality may occur under such conditions. However, if a person gets the necessary organ under such conditions, its life can be saved.

Availability of donor is an important requirement in organ transplantation. Each person has a pair of kidneys. As the process of excretion can occur with the help of single kidney, person can donate another one. Similarly, skin from certain parts of the body can also be donated.

Various factors like blood group, diseases, disorders, age, etc. of the donor and recipient need to be paid attention during transplantation.

However, other organs cannot be donated during life time. Organs like liver, heart, eyes can be donated after death only. This has lead to the emergence of concepts like posthumous (after death) donation of body and organs.

Organ and Body Donation: human bodies are disposed off after death as per traditional customs. However due to progress in science, it has been realized that many organs remain functional for certain period even after death occurs under specific conditions. Concepts like organ donation and body donation have emerged recently after realization that such organs can be used to save the life of other needful persons. A liberal view behind the concept of organ and body donation is that after death, our body should be useful to other needful persons so that their miserable life would become comfortable. Awareness about these concepts is increasing in our country and people are voluntarily donating their bodies.

Life of many people can be saved by organ and body donation. Blinds can regain the vision. Life of many people can be rendered comfortable by donation of organs like liver, kidneys, heart, heart valves, skin. etc. Similarly, body can be made available for research in medical studies. Many government and social organizations are working towards increasing the awareness about body donation.

Question 5.
Which fruits processing industries you observe in your surrounding? What is their effect? (Make a list and discuss: Textbook page no. 99)
Answer:

Fruit Processing:
we are daily using various products prepared from fruits. All are consuming the products like chocolates, juices, jams and jellies. All these products can be produced by processing on fruits. Fruits are perishable agro-produce. It needs the processing in such a way that it can be used throughout the year. Fruit processing includes various methods ranging from storage in cold storage to drying, salting, air tight pucking, preparing murabba, evaporating, etc.

Projects: (Do it your self)

Project 1.
Collect information about various hybrid varieties of animals. What are their benefits? Make a presentation of various pictures and videos. (Use of ICT: Textbook page no. 93)

Project 2.
Visit the websites: http://www.who.int/transplantation/organ/en/ and www.organindia.org / approaching-the- transplant/and collect more information about ‘brain dead’, organ donation and body donation (Internet is my friend: Textbook page no. 90)

Project 3.
Collect more information about the Human Genome Project, one of the important projects in the world.
(Internet is my friend: Textbook page no. 95)

Project 4.
Collect the information and make the chart about the work of various state and national-level institutes related with biotechnology. (Internet is my friend: Textbook page no. 97)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Miscellaneous Exercise 6

Question 1.
Find the value of r if ⁵⁶C_r+6 : ⁵⁴P_r-1 = 30800 : 1
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R
The number of words starting with A = 5!
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
The number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
Find the number of ways of distributing n balls in n cells. What will be the number of ways if each cell must be occupied?
Solution:
There are n balls and n cells
(i) Every ball can be put in any of the n cells.
Number of distributions = n × n × …… × n = (n)ⁿ
(ii) For filling the first cell, n balls are available.
The first cell is filled in n ways.
The second cell is filled in (n – 1) ways
The third cell is filled in (n – 2) ways and so on.
the nth cell is tilled in one way.
Required number = n(n – 1)(n – 2) …… 1 = n!

Question 5.
Thane is the 20th station from C.S.T. If a passenger can purchase a ticket from any station to any other station, how many different tickets must be available at the booking window?
Solution:
Taking CST as the first station and Thane as 20th,
Let us name CST as A₀ next station as A₁ and so on, Thane is A₂₀
From station A₀, 20 different journeys are possible
From station A₁, 20 different journeys are possible.
From station A₂₀, 20 different journeys are possible.
Total number of different tickets of different journeys = 21 × 20 = 420

Question 6.
English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
Number of 3 Letter passwords = ¹¹P₃
= 11 × 10 × 9
= 990

Question 7.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also among the given numbers 2 repeats twice and 3 repeats thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= \(\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}\)
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360
∴ 360 numbers that exceed one million can be formed with the digits 3, 2, 0, 4, 3, 2, 3.

Question 8.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:

Required number = ²⁶C₆ + ²⁶C₁₀

Question 9.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:

Required Number = 5 + 10 = 15

Question 10.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so. Solution:
Required number = ³⁰C₇ × ²³C₁₀ × ¹³C₁₃

Question 11.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 2⁷ = 128
This number includes one case when the student passes in all subjects.
∴ Required number = 128 – 1 = 127

Question 12.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:

Question 13.
Five balls are to be placed in three boxes, where each box can contain upto five balls. Find the number of ways if no box is to remain empty.
Solution:
Let boxes be named as I, II, III
Let sets A, B, C represent cases in which boxes I, II, III remain empty
Then A ∪ B ∪ C represent the cases in which at least one box remains empty.
Then we use method of indirect counting
Required number = Total number of distributions – n(A ∪ B ∪ C) …..(i)
n(A ∪ B ∪ C) represent the number of undesirable cases
Total number of distributions = 3 × 3 × 3 × 3 × 3 = 3⁵ = 243 …….(ii)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C) …..(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
∴ n(A) + n(B) + n(C) = 3 × 2⁵ ……(iv)
If boxes I and II remain empty then all balls go to box III
Similarly we would have two more cases.
∴ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 3 × 1⁵ ……(v)
∴ n(A ∩ B ∩ C) = 0 …….(vi) [as all boxes cannot be empty]
Substitute from (iv), (v), (vi) to (iii) to get
n(A ∪ B ∪ C) = 3 × 2⁵ – 3 × 1⁵
= 96 – 3
= 93
Substitute n(A ∪ B ∪ C) and from (ii) to (i), we get
Required number = 243 – 93 = 150

Question 14.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 2¹²
This number includes the case in which all 12 lamps are OFF.
∴ Required Number = 2¹² – 1 = 4095

Question 15.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
Let the quadratic equation be ax² + bx + c = 0, a ≠ 0

∴ Required number = 3 × 4 × 4 = 48

Question 16.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
Let the telephone number be 45abcd

∴ Required number = ⁸P₄ = 1680

Question 17.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 question.
Number of outcomes = 2⁶
This number includes the case when the student solves NONE of the question.
Required number = 2⁶ – 1
= 64 – 1
= 63

Question 18.
Find the number of ways of dividing 20 objects in three groups of sizes 8, 7 and 5.
Solution:
Select 8 objects out of 20 in ²⁰C₈ ways
Select 7 objects from the remaining 12 in ¹²C₇ ways and 5 objects from the remaining 5 in ⁵C₅ ways
Required number is = ²⁰C₈ × ¹²C₇ × ⁵C₅

Question 19.
There are 8 doctors and 4 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 8 doctors and 4 lawyers.
We need to select a team of 6 which contains at least one doctor.
Since there are only 4 lawyers any team of 6 will contain at least two doctors.
Required number = ¹²C₆ = 924

Question 20.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms that can be formed.
Solution:
We need 2 lines from each set.
Required number = ⁴C₂ × ⁵C₂
= 6 × 10
= 60

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.7 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.7

Question 1.
Find n if ⁿC₈ = ⁿC₁₂
Solution:
ⁿC₈ = ⁿC₁₂
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20

Question 2.
Find n if ²³C_3n = ²³C_2n+3
Solution:
²³C_3n = ²³C_2n+3
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

Question 3.
Find n if ²¹C_6n = \({ }^{21} C_{n^{2}+5}\)
Solution:
²¹C_6n = \({ }^{21} C_{n^{2}+5}\)
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 6n = n² + 5 or 6n = 21 – (n² + 5)
∴ n² – 6n + 5 = 0 or 6n = 21 – n² – 5
∴ n² – 6n + 5 = 0 or n² + 6n – 16 = 0
If n² – 6n + 5 = 0 then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then n² + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n² + 6n – 16 = 0 then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2

Question 4.
Find n if ²ⁿC_r-1 = ²ⁿC_r+1
Solution:
²ⁿC_r-1 = ²ⁿC_r+1
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Question 5.
Find n if ⁿC_n-2 = 15
Solution:
ⁿC_n-2 = 15
∴ ⁿC₂ = 15 …….[∵ ⁿC_r = ⁿC_n-r]
∴ \(\frac{n !}{(n-2) ! 2 !}\) = 15
∴ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) ! 2 \times 1}\) = 15
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Comparing both sides, we get
∴ n = 6

Question 6.
Find x if ⁿP_r = x ⁿC_r
Solution:

Question 7.
Find r if ¹¹C₄ + ¹¹C₅ + ¹²C₆ + ¹³C₇ = ¹⁴C_r
Solution:

Question 8.
Find the value of \(\sum_{r=1}^{4}{ }^{21-r} C_{4}+{ }^{17} C_{5}\)
Solution:

Question 9.
Find the differences between the largest values in the following:
(i) ¹⁴C_r – ¹²C_r
Solution:
Greatest value of ¹⁴C_r
Here n = 14, which is even
Greatest value of ⁿC_r occurs at r = \(\frac{n}{2}\) if n is even

∴ Difference between the greatest values of ¹⁴C_r and ¹²C_r = 3432 – 924 = 2508

(ii) ¹³C_r – ⁸C_r
Solution:
Greatest value of ¹³C_r
Here n = 13, which is odd
Greatest value of ⁿC_r occurs at r = \(\frac{\mathrm{n}-1}{2}\) if n is odd

∴ Difference between the greatest values of ¹³C_r and ⁸C_r = 1716 – 70 = 1646

(iii) ¹⁵C_r – ¹¹C_r
Solution:
Greatest value of ¹⁵C_r
Here n = 15, which is odd
Greatest value of ⁿC_r occurs at r = \(\frac{\mathrm{n}-1}{2}\) if n is odd

Difference between the greatest values of ¹⁵C_r and ¹¹C_r = 6435 – 462 = 5973

Question 10.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more join the party = 10 + 5 + 1 = 16

Question 11.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consists of at least 3 women.

∴ Number of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.

Question 12.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in the majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women
Consider the following table:


∴ Number of committees = 14112 + 14700 + 6720 + 1260 + 81 = 36873
∴ At least 5 women are there in 36873 committees.

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= ¹⁸C₁₀ – 36873
= ¹⁸C₈ – 36873
= 43758 – 36873
= 6885

Question 13.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

Question 14.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicketkeeper and at least 4 bowlers are to be included in the team.
Consider the following table:

∴ Number of ways a team of 11 players can be selected = 45045 + 6006 = 51051

Question 15.
Five students are selected from 11. How many ways can these students be selected if
(i) two specified students are selected?
(ii) two specified students are not selected?
Solution:
5 students are to be selected from 11 students
(i) When 2 specified students are included
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = ⁹C₃
= \(\frac{9 !}{3 ! \times 6 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\)
= 84
∴ Selection of students is done in 126 ways when 2 specified students are not selected.

(ii) When 2 specified students are not included then 5 students can be selected from the remaining (11 – 2) = 9 students
∴ Number of ways of selecting 5 students from 9 students = ⁹C₅
= \(\frac{9 !}{5 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}\)
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not selected.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.6 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.6

Question 1.
Find the value of
(i) ¹⁵C₄
Solution:

(ii) ⁸⁰C₂
Solution:

(iii) ¹⁵C₄ + ¹⁵C₅
Solution:

(iv) ²⁰C₁₆ – ¹⁹C₁₆
Solution:

Question 2.
Find n if
(i) ⁶P₂ = n ⁶C₂
Solution:

(ii) ²ⁿC₃ : ⁿC₂ = 52 : 3
Solution:

(iii) ⁿC_n-3 = 84
Solution:
ⁿC_n-3 = 84
∴ \(\frac{n !}{(n-3) ![n-(n-3)] !}\) = 84
∴ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{(\mathrm{n}-3) ! \times 3 !}\) = 84
∴ n(n – 1) (n – 2) = 84 × 6
∴ n(n – 1) (n – 2) = 9 × 8 × 7
Comparing on both sides, we get
∴ n = 9

Question 3.
Find r if ¹⁴C_2r : ¹⁰C_2r-4 = 143 : 10
Solution:

∴ \(\frac{14 \times 13 \times 12 \times 11}{2 \mathrm{r}(2 \mathrm{r}-1) \times(2 \mathrm{r}-2)(2 \mathrm{r}-3)}=\frac{143}{10}\)
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if.
(i) ⁿP_r = 720 and ⁿC_n-r = 120
Solution:

(ii) ⁿC_r-1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42
Solution:

Question 5.
If ⁿP_r = 1814400 and ⁿC_r = 45, find r.
Solution:

∴ r! = 40320
∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
∴ r! = 8!
∴ r = 8

Question 6.
If ⁿC_r-1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 green balls and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in ⁶C₃ ways.
3 green balls can be selected from 5 green balls in ⁵C₃ ways.
3 blue balls can be selected from 7 blue balls in ⁷C₃ ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in ⁶C₃ ways.
2 girls can be selected from 4 girls in ⁴C₂ ways.
∴ Number of ways the team can be selected = ⁶C₃ × ⁴C₂
= \(\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{4 \times 3 \times 2 !}{2 \times 2 !}\)
= 20 × 6
= 120
∴ The team of 3 boys and 2 girls can be selected in 120 ways.

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = ⁿC₂
Given 66 handshakes were exchanged.
∴ 66 = ⁿC₂
∴ 66 = \(\frac{n !}{2 !(n-2) !}\)
∴ 66 × 2 = \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) !}\)
∴ 132 = n(n – 1)
∴ n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides. .
∴ Through ‘n’ points we can draw ⁿC₂ lines including sides.
∴ Number of diagonals in n sided polygon = ⁿC₂ – n (∴ n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel
∴ they intersect at a point
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 10 points on a plane.
(i) No three of them are collinear:
Since a line is obtained by joining 2 points,
number of lines passing through these points if no three points are collinear = ¹⁰C₂
= \(\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)
= 5 × 9
= 45

(ii) When 4 of them arc collinear:
∴ Number of lines passing through these points if 4 points are collinear
= ¹⁰C₂ – ⁴C₂ + 1
= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1
= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 12 points on the plane
(i) When no three of them are collinear:
Since a triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = ¹²C₃
= \(\frac{12 !}{3 ! 9 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)
= 220

(ii) When 4 of these points are collinear:
∴ Number of triangles that can be obtained from these points = ¹²C₃ – ⁴C₃
= 220 – \(\frac{4 !}{3 ! \times 1 !}\)
= 220 – \(\frac{4 \times 3 !}{3 !}\)
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
Out of 8 consonants, 4 can be selected in ⁸C₄
= \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)
= 70 ways
From 3 vowels, 2 can be selected in ³C₂
= \(\frac{3 !}{2 ! 1 !}\)
= \(\frac{3 \times 2 !}{2 !}\)
= 3 ways
Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in ⁶P₆ i.e., 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.5

Question 1.
In how many different ways can 8 friends sit around a table?
Solution:
We know that ‘n’ persons can sit around a table in (n – 1)! ways
∴ 8 friends can sit around a table in 7! ways
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.
∴ 8 friends can sit around a table in 5040 ways.

Question 2.
A party has 20 participants and a host. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Solution:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 – 1)! = 20! ways.
When two particular participants be seated on either side of the host.
Host takes chair in 1 way.
These 2 persons can sit on either side of host in 2! ways
Once host occupies his chair, it is not circular permutation any more.
Remaining 18 people occupy their chairs in 18! ways.
∴ Total number of arrangement possible if two particular participants be seated on either side of the host = 2! × 18!

Question 3.
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are
(i) always together.
(ii) never together.
Solution:
(i) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit.
They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total 23) which can be done in (23 – 1)! = 22! ways.
∴ The total number of arrangements if two specified delegates are always together = 22! × 2!

(ii) When 2 specified delegates are never together then, the other 22 delegates can participate in a round table discussion in (22 – 1)! = 21! ways.
∴ There are 22 places of which any 2 places can be filled by those 2 delegates who are never together.
∴ Two specified delegates can be arranged in ²²P₂ ways.
∴ Total number of arrangements if two specified delegates are never together = ²²P₂ × 21!
= \(\frac{22 !}{(22-2) !}\) × 21!
= \(\frac{22 !}{20 !}\) × 21!
= 22 × 21 × 21!
= 21 × 22 × 21!
= 21 × 22!

Question 4.
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Solution:
There are 15 people to sit around a table.
∴ They can be arranged in (15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e. clockwise and anticlockwise) coincide.

∴ Number of arrangements possible for not to have same neighbours = \(\frac{14 !}{2}\)

Question 5.
A committee of 20 members sits around a table. Find the number of arrangements that have the president and the vice president together.
Solution:
A committee of 20 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 18 members of the committee is to be arranged around a table, which can be done in (19 – 1)! = 18! ways.
∴ The total number of arrangements possible if President and Vice-president sit together = 18! × 2!

Question 6.
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated
(i) between the two women.
(ii) between two men.
Solution:
(i) 5 men, 2 women, and a child sit around a table
When a child is seated between two women
∴ The two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Also, these 3 are to be seated with 5 men, (i.e. a total of 6 units) which can be done in (6 – 1)! = 5! ways.
∴ The total number of arrangements if the child is seated between two women = 5! × 2!

(ii) Two men out of 5 men can sit on either side of the child in 5P2 ways.
Let us take two men and a child as one unit.
Now these are to be arranged with the remaining 3 men and 2 women
i.e., a total of 6 events (3 + 2 + 1) is to be arranged around a round table which can be done in (6 – 1)! = 5! ways.
∴ The total number of arrangements, if the child is seated between two men = ⁵P₂ × 5!

Question 7.
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Solution:
8 men can be seated around a table in (8 – 1)! = 7! ways.
There are 8 gaps created by 8 men’s seats.
∴ 6 Women can be seated in 8 gaps in ⁸P₆ ways
∴ Total number of arrangements so that no two women are together = 7! × ⁸P₆

Question 8.
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Solution:

Two women sit together and one woman sits separately.
Women sitting separately can be selected in 3 ways.
The other two women occupy two chairs in one way (as it is a circular arrangement).
They can be seated on those two chairs in 2 ways. Suppose two chairs are chairs 1 and 2 shown in the figure.
Then the third woman has only two options viz chairs 4 or 5.
∴ The third woman can be seated in 2 ways. 3 men are seated in 3! ways
∴ Required number = 3 × 2 × 2 × 3!
= 12 × 6
= 72

Question 9.
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Solution:
Ten things can be arranged in a circular order of which 4 are alike in \(\frac{9 !}{4 !}\) ways.
∴ Required total number of arrangements = \(\frac{9 !}{4 !}\)

Question 10.
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Solution:
Since 2 particular persons can’t be sitting side by side.
The other 13 persons can be arranged around the table in (13 – 1)! = 12!
13 people around a table create 13 gaps in which 2 people are to be seated
Number of arrangements of 2 people = ¹³P₂
∴ The total number of arrangements in which two specified persons not sitting side by side = 12! × ¹³P₂
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 1.
a. Alloy of sodium with mercury.
Answer:
Silver amalgam.

b.Molecular formula of common ore of aluminium.
Answer:
Al₂O₃.nH₂O

c. The oxide that forms salt and water by reacting with both acid and base.
Answer:
Aluminium oxide (Al₂O₃).

d. device used for grinding an ore.
Answer:
The device used for grinding an ore is grinding mill.

e. The nonmetal having electrical conductivity.
Answer:
Graphite having electrical conductivity.

f. The reagent that dissolves noble metals.
Answer:
Aqua regia is the reagent that dissolves noble metals like gold and platinum.

Question 2.
Make pairs of substances and their properties.

Answer:
(1) Potassium bromide – Soluble in water
(2) Gold – High ductility
(3) Sulphur – Combustible
(4) Neon – No chemical reaction

Question 3.
Identify the pairs of metals and their ores from the following.

Answer:
(1) Bauxite – Aluminium
(2) Cassiterite – Tin
(3) Cinnabar – Mercury

Question 4.
Explain the terms.
a. Metallurgy
Answer:
Metallurgy: The process used for extraction of metals in their pure form from their ores, then metals are further purified by different methods of purification. All the process is called metallurgy.

b. Ores.
Answer:
Ores: The minerals from which metals are extracted profitably and conveniently are called ores.
Examples: Bauxite (Al₂O₃.H₂O), Cinnabar (HgS).

c. Minerals.
Answer:
Minerals: The naturally occurring compounds of metals along with other impurities are known as minerals.
Examples: Rocks are composed of mixtures of minerals. Talc and granite are minerals.

d. Gangue.
Answer:
Gangue: Ores contain metal compounds with some of the impurities like soil, sand, rocky material, etc. These impurities are called gangue.

Question 5.
Write scientific reasons.
a. Lemon or tamarind is used for cleaning copper vessels turned greenish.
Answer:

• Copper undergoes oxidation in air to form black copper oxide. Copper oxide reacts slowly with carbon dioxide in air and gains a green coat. This green substance is copper carbonate.
• Lemon and tamarind contain acid. The acid dissolves the green coating of basic copper carbonate present on the surface of a tarnished copper utensil and makes it shiny again.

b. Generally the ionic compounds have high melting points.
Answer:

• The ionic compounds exist in solid state and are hard due to strong electrostatic force of attraction between oppositely charged ions.
• The intermolecular force of attraction is high in ionic compounds and large energy is required to overcome it. Therefore, ionic compounds have high melting points.

c. Sodium is always kept in kerosene.
(OR)
Why is sodium stored in kerosene?
Answer:

• Sodium reacts so vigorously with atmospheric oxygen that it catches fire if kept in the open.
• It does not react with kerosene and sinks in it. Hence, to protect sodium and to prevent accidental fires it is always kept in kerosene.

d. Pine oil is used in the froth floatation process.
Answer:

• In the concentration of an ore by froth floatation process, the ore is mixed with water and pine oil. When air is bubbled through the mixture a froth is formed.
• The mineral particles in the ore are preferentially wetted by the oil and float on the top in the froth.
• The gangue particles are wetted by water and settle down. Thus the mineral can be separated from the gangue and the ore is concentrated.

e. Anodes need to be replaced from time to time during the electrolysis of alumina.
Answer:

• During electrolysis of alumina, the oxygen liberated at the carbon anode reacts with graphite rods (carbon anode) and forms carbon dioxide.
• As the anodes get oxidised during electrolysis of alumina, they are continuously eroded. Hence, it is necessary to replace anodes from time to time.

Question 6.
When a copper coin is dipped in silver nitrate solution, a glitter appears on the coin after some time. Why does this happen? Write the chemical equation.
Answer:
When a copper coin is dipped in a silver nitrate solution, more reactive copper displaces silver from silver nitrate solution. The silver so liberated deposits on the copper coin. As a result, a shiny coat of silver is formed on the coin.
Cu_(s) + 2AgNO_3(aq) → Cu(NO₃)_2(aq) + 2Ag_(s)

Question 7.
The electronic configuration of metal ‘A’ is 2, 8, 1 and that of metal ‘B’ is 2, 8, 2. Which of the two metals is more reactive? Identify these metals. Write their reaction with dilute hydrochloric acid. (Practice Activity Sheet – 1)
Answer:
If the number of electrons in the outermost orbit is less, then the metal is more reactive. Metal A contains one electron in the outermost shell, while metal B contains two electrons. Hence, metal A is more reactive than metal B.

Metal A is sodium and metal B is magnesium. Reactions of Na and Mg with dil. HCl are,

Question 8.
Draw a neat labelled diagram.
a. Magnetic separation method.
Answer:

b. Forth floatation.
Answer:

c. Electrolytic reduction of alumina.
Answer:

d. Hydraulic separation method.
Answer:

Question 9.
Write chemical equation for the following events.
a. Aluminium came in contact with air.
Answer:
When aluminium is exposed to air, it develops a thin oxide layer of aluminium.

b. Iron filings are dropped in aqueous solution of copper sulphate.
Answer:
When iron filings are dropped in copper sulphate solution, more reactive iron displaces copper from copper sulphate solution. The iron filings get coated with reddish brown copper metal and the blue colour of copper sulphate fades gradually and ferrous sulphate is formed.

c. A reaction was brought about between ferric oxide and aluminium.
Answer:
The reaction between ferric oxide and iron produces aluminium oxide and iron. It is a thermite reaction and is highly exothermic.
It produces a large amount of heat, which is released to melt oxygen and aluminium. This reaction is used in welding of machineries. It is also used in warfare to make grenades.

The chemical reaction for the above is as follows:
3Fe₃O₂ + 4Al → 2Al₂O₃ + 6Fe

d. Electrolysis of alumina is done.
Answer:
During electrolysis of alumina, aluminium is deposited at the cathode. Molten aluminium being heavier than the electrolyte, is collected at the bottom of the tank. Oxygen gas is liberated at the anode.
Anode reaction: 2O^–– → O₂ + 4e^– (Oxidation)
Cathode reaction: Al³⁺ + 3e^– → Al_(l) (Reduction)

e. Zinc oxide is dissolved in dilute hydrochloric acid.
Answer:
Zinc oxide is dissolved in dilute hydrochloric acid, zinc chloride and water are formed.

Question 10.
Complete the following statement using every given options.
During the extraction of aluminium
a. Ingredients and gangue in bauxite.
b. Use of leuching during the concentration of ore.
c. Chemical reaction of transformation of bauxite into alumina by Hall’s process.
d. Heating the aluminium ore with concentrated caustic soda.
Answer:
c. Chemical reaction of transformation of bauxite into alumina by Hall’s process.

Question 11.
Divide the metals Cu, Zn, Ca, Mg, Fe, Na, Li into three groups, namely, reactive metals, moderately reactive metals and less reactive metals.
Answer:
Reactive metals: Na, Li, Ca
Moderately reactive metals: Zn, Fe, Mg,
Less reactive metals: Cu

Project: (Do it your self)

Collect metal vessels and various metal articles. Write detailed informciton. write the steps in the procedure that can be done in the laboratory for giving glitter to these. Seek guidance from your teacher.

Can you recall? (Text Book Page No. 93)

Question 1.
what are the physical properties of metals and nonmetals?
Answer:
Properties of metals:

1. Solid state (Exception: Mercury and gallium)
2. Typical lustre
3. Malleability and ductility
4. Hardness (Exception: Lithium, sodium and potassium)
5. Good conductors of heat and electricity
6. High melting and boiling points (On the other hand, the melting and boiling points of the metals sodium, potassium, mercury and galium are very low.)
7. Sonorous and produce sound on striking a hard surface.

Properties or nonmetals:

1. Gaseous or solid state (Exception: Bromine in liquid state)
2. Lack of any typical lustre (Exception: Iodine and Diamond)
3. Brittleness in the solid state (Exception: Diamond is the hardest natural substance)
4. Bad conductors of heat and electricity (Exception: Graphite) (Diamond is good conductor of heat)
5. Low melting and boiling points.

Can you recall? (Text Book Page No. 102)

Question 1.
What is the electronic definition of oxidation and reduction?
Answer:
When a metal loses çlectrons the process is called an oxidation while when a nonmetal gains electrons, it is called a reduction,
Na → Na⁺ + e^– (oxidation)
Cl^– + e^– → Cl (reduction)

Can you recall? (Text Book Page No. 106)

Question 1.
What is meant by corrosion?
Answer:
Corrosion is degradation of a material due toreaction with its environment.

Question 2.
Have you observed the following things?
(1) Old iron bars in the builthngs.
Answer:
When old iron bars in the buildings are exposed to moist air for a long time, they acquire a coating of browm nlaky substance called rust. (Fe₂O₃.H₂O)

(2) Copper vessels not cleaned for a long time.
Answer:
If copper vessels are not cleaned for a long time, they react with moist carbon dioxide in the air, lose their shine and gain a green coat of copper carbonate. (CuCO3)

Question 3.
Silver ornaments or idols exposed to air for a long time.
Answer:
When silver ornaments or idols are kept exposed to air for a long time, silver reacts with sulphur in the air to form a coating of black silver sulphide. (Ag₂S)

Question 4.
Old vehicles fit to be thrown away.
Answer:
The metallic parts of the body of old cars are corroded, eaten up and sometimes become perforated. The old cars also lose the original colour due to formation of flakes of rust.

Use your brain power! (Text Book Page No. 98)

Question 1.
In the reaction between chlorine and HBr a transformation or HBr into Br₂ takes place. Can this transformation be called oxidation? What Is the oxidant that brings about this oxidation?
Answer:
The conversion of HBr to Br₂ is an oxidation process. In the above reaction, Cl₂ in the oxidant.

Can you tell? (Text Book Page No. 104)

Question 1.
what are the moderately reactive metals?
Answer:
In the middle of the reactivity series, metals such as iron. zinc, lead, copper are moderately reactive.

Question 2.
In which form to the moderately reactive metals occur in nature?
Answer:
The moderately reactive metals which occur in nature are in the form of their sulphide salts or carbonate salts.

Think about it (Text Book Page No. 106)

Question 1.
Why do silver articles turn blackish while copper vessels turn greenish on keeping in air for long time?
Answer:

1. Silver articles turn blackish on exposure to air for a long time. This is because of silver sulphide (Ag₂S) laver formed on the silver articles by the reaction of silver with hydrogen sulphide.
2. Carbon dioxide in moist air reacts with copper vessel. Copper loses its lustre due to formation of greenish layer of copper carbonate (CuCO₃) on its surface.

Question 2.
Why do pure gold and platinum always glitter?
Answer:
Gold and platinum are noble metals as they do not react with moisture, O₂ and CO₂ from air also acids and alkalis, therefore, pure gold and platinum always glitter.

Use your brain power! (Text Book Page No. 103)

Question 1.
Write the electrode reaction for electrolysis of molten magnesium chloride and calcium chloride.
Answer:
(1) Magnesium chloride (MgCl₂):
MgCl₂ → Mg²⁺ + 2Cl^–
At the cathode: Mg²⁺ + 2e^– → Mg
At the anode: 2Cl^– → Cl₂ + 2e^–

(2) Calcium chloride (CaCl2):
At the cathode: Ca²⁺ + 2e^– → Ca
At the anode: 2Cl^– → Cl₂ + 2e^–

Can you tell? (Text Book Page No. 106)

Question 1.
Which measures would you suggest to stop the corrosion of metallic articles or not allow the corrosion to start?
Answer:
Various types of methods are used to protect metals from corrosion. Almost in all the methods, special attention is paid so that iron does not rust. It is possible to lower the rate of the process of rusting of iron. Corrosion of metals can be stopped by detaching the air from metals.

Some methods are as follows :

1. To fix a layer of some substance on the metal surface so that the contact of the metal with moisture and oxygen in the air is prevented and no reaction would occur between them.
2. To prevent corrosion of metals by applying a layer of paint, oil, grease or varnish on their surface. For example, corrosion of iron can be prevented by this method.

Question 2.
What is done so to prevent rusting of iron windows and iron doors of your house?
Answer:
To prevent rusting of iron windows and iron doors in the house, they are painted so that they do not rust.

Question 3.
What is done so to prevent rusting or iron windows and iron doors of your house?
Answer:
To prevent rusting of iron windows and iron doors in the house, they are painted so that they do not rust.

Use your brain power! (Text Book Page No. 107)

Question 1.
Can we permanently prevent the rusting of an iron article by applying a layer of paint on its surface?
Answer:
The method of painting is alright for some time. We cannot protect the articles permanently from rusting by painting them.

Question 2.
Why do new iron sheets appear shiny?
Answer:
The new iron sheets appear shiny because a layer of non-corrosionable metal is fixed on the surface of corrosionable metal.

Collect information. (Text Book Page No. 108)

Question 1.
What are the various alloys used in daily life? Where are those used?
Answer:

Question 2.
What are the properties that the alloy used for minting coins should have?
Answer:
The alloy used for minting coins should have excellent wear resistance and anti-corrosion properties.

Fill in the blanks:

Question 1.
……………has the highest melting point.
Answer:
Tungsten has the highest melting point.

Question 2.
Mercury and…………are two metals in the liquid state at room temperature.
Answer:
Mercury and galium are two metals in the liquid state at room temperature.

Question 3.
…………is the hardest natural substance.
Answer:
Diamond is the hardest natural substance.

Question 4.
The naturally occurring compounds of metals along with other impurities are known as………….
Answer:
The naturally occurring compounds of metals along with other impurities are known as minerals.

Question 5.
The minerals from which metals are extracted profitably and conveniently are called…………..
Answer:
The minerals from which metals are extracted profitably and conveniently are called ores.

Question 6.
An ore contains some of the impurities like soil, sand, etc. These impurities are called…………
Answer:
An ore contains some of the impurities like soil, sand, etc. These impurities are called gangue.

Question 7.
The process of extraction of a metal from its ore is called……….
Answer:
The process of extraction of a metal from its ore is called metallurgy.

Question 8.
Bauxite is a common ore of………..
Answer:
Bauxite is a common ore of aluminium.

Question 9.
…………. process is used for the purification of bauxite.
Answer:
Bayer’s process is used for the purification of bauxite.

Question 10.
During the electrolysis of alumina, ………..is liberated at the anode.
Answer:
During the electrolysis of alumina, oxygen is liberated at the anode.

Question 11.
The reaction of iron oxide with aluminium is known as…………..reaction.
Answer:
The reaction of iron oxide with aluminium is known as thermit reaction.

Question 12.
The process of coating a thin layer of zinc on iron is known as…………
Answer:
The process of coating a thin layer of zinc on iron is known as galvanising.

Question 13.
The metal that produces a sound on striking a hard surface is said to be………….
Answer:
The metal that produces a sound on striking a hard surface is said to be sonorous.

Question 14.
The process in which carbonate ores are changed into oxides by heating strongly in limited air is known as …………….
Answer:
The process in which carbonate ores are changed into oxides by heating strongly in limited air is known as calcination.

Question 15.
…………compounds are insoluble in solvents like kerosene and petrol.
Answer:
Ionic compounds are insoluble in solvents like kerosene and petrol.

Question 16.
…………… is used to obtain pure metals from impure metals.
Answer:
Electrolys is method is used to obtain pure metals from impure metals.

Question 17.
Corrosion can be prevented-by putting a layer of…………metal on corrosionable metal.
Answer:
Corrosion can be prevented by putting a layer of non-corrosionable metal on corrosionable metal.

Rewrite the following statements by selecting the correct options:

Question 1.
………… is a metal.
(a) Mg
(b) S
(c) P
(d) Br
Answer:
Mg is a metal.

Question 2.
………. is a nonmetal.
(a) Au
(b) Hg
(c) Br
(d) Cu
Answer:
Br is a nonmetal.

Question 3.
………… is a metalloid.
(a) Aluminium
(b) Antimony
(c) Zinc
(d) Mercury
Answer:
Antimony is a metalloid.

Question 4.
Metalloids have properties of ………..
(a) metals
(b) nonmetals
(c) both metals and nonmetals
(d) neither metals nor nonmetals
Answer:
Metalloids have properties of both metals and nonmetals.

Question 5.
…………. is a good conductor of electricity.
(a) Bromine
(b) Iodine
(c) Graphite
(d) Sulphur
Answer:
Graphite is a good conductor of electricity.

Question 6.
………. is a metal which is in liquid form at ordinary temperature and pressure.
(a) Magnesium
(b) Sodium
(c) Scandium
(d) Mercury
Answer:
Mercury is a metal which is in liquid form at ordinary temperature and pressure.

Question 7.
………. is an amphoteric oxide.
(a) Na₂O
(b) MgO
(c) ZnO
(d) SO₂
Answer:
ZnO is an amphoteric oxide.

Question 8.
……….. is an acidic oxide.
(a) Na₂O
(b) CO₂
(c) FeO₃
(d) H₂O
Answer:
CO₂ is an acidic oxide.

Question 9.
………. is a basic oxide.
(a) CO₂
(b) K₂O
(C) SO₂
(d) Al₂O₃
Answer:
K₂O is a basic oxide.

Question 10.
………… is an ore of aluminium.
(a) Cryolite
(b) Bauxite
(c) Haematite
(d) Aluminium carbonate
Answer:
Bauxite is an ore of aluminium.

Question 11.
Bronze is an alloy of ………..
(a) copper and tin
(b) copper and zinc
(c) copper and iron
(d) iron and nickel
Answer:
Bronze is an alloy of copper and tin.

Question 12.
An alloy prepared from iron, nickel and chromium is known as …………
(a) brass
(b) bronze
(c) stainless steel
(d) amalgam
Answer:
An alloy prepared from iron, nickel and chromium is known as stainless steel.

Question 13.
…………. is an allotropic form of a nonmetal which conducts electricity.
(a) Sulphur
(b) Graphite
(c) Chlorine
(d) Iodine
Answer:
Graphite is an allotropic form of a nonmetal which conducts electricity.

Question 14.
………….. has an oxide which is soluble in sodium hydroxide.
(a) Calcium
(b) Magnesium
(c) Iron
(d) Zinc.
Answer:
Zinc has an oxide which is soluble in sodium hydroxide.

Question 15.
………… prevents the rusting of iron.
(a) Copper
(b) Zinc
(c) Aluminium
(d) Silver
Answer:
Zinc prevents the rusting of iron.

Question 16.
………….. is obtained by the reduction of its oxide by carbon.
(a) Zinc
(b) Aluminium
(c) Sodium
(d) Potassium
Answer:
Zinc is obtained by the reduction of its oxide by carbon.

Question 17.
………….. is used as an anode during the electrolytic reduction of bauxite.
(a) Sulphur
(b) Graphite
(c) Platinum
(d) Aluminium
Answer:
Graphite is used as an anode during the electrolytic reduction of bauxite.

Question 18.
Silver gets corroded due to ………… in air.
(a) oxygen
(b) hydrogen sulphide
(c) carbon dioxide
(d) nitrogen
Answer:
Silver gets corroded due to hydrogen sulphide in air.

Question 19.
…………. is the hardest substance and has the highest melting and boiling points.
(a) Iodine
(b) Sulphur
(c) Diamond
(d) Phosphorus
Answer:
Diamond is the hardest substance and has the highest melting and boiling points.

Question 20.
Jewellery articles are gold plated ………….
(a) to prevent corrosion
(b) to prevent rusting of the base metal
(c) to make articles attractive
(d) all of these
Answer:
(d) all of these

Question 21.
To show that zinc is more reactive than copper, the correct procedure is to ………..
(a) prepare copper sulphate solution and dip a zinc strip in it
(b) prepare zinc sulphate solution and dip a copper strip in it
(c) heat together zinc and copper strips
(d) add dil. nitric acid to both the strips
Answer:
To show that zinc is more reactive than copper, the correct procedure is to prepare copper sulphate solution and dip a zinc strip in it.

Question 22.
Iron is ………
(a) more reactive than zinc
(b) more reactive than aluminuium
(c) less reactive than copper
(d) less reactive than aluminium
Answer:
Iron is less reactive than aluminium.

Question 23.
A solution of Al₂(SO₄)₃ in water is …………
(a) blue
(b) pink
(c) green
(d) colourless
Answer:
A solution of Al₂(SO₄)₃ in water is colourless.

Question 24.
A solution of ………… in water is blue in colour.
(a) CuSO₄
(b) FeSO₄
(c) ZnSO₄
(d) Al₂(SO₄)₃
Answer:
A solution of CuSO₄ in water is blue in colour.

Question 25.
A solution of …………. n water is green in colour.
(a) CuSO₄
(b) FeSO₄
(c) ZnSO₄
(d) Al₂(SO₄)₃
Answer:
A solution of FeSO₄ in water is green in colour.

Question 26.
What would be the correct order if Zn, Fe, Al and Cu are arranged in increasing order of reactivity?
(a) Cu, Fe, Zn, Al
(b) Al, Cu, Fe, Zn
(c) Zn, Al, Cu, Fe
(d) Fe, Zn, Al, Cu (Practice Activity Sheet – 2)
Answer:
(a) Cu, Fe, Zn, Al

Question 27.
During the extraction of aluminium ……….
(a) Ingredients and gangue in bauxite
(b) Use of leaching during the concentration of ore
(c) Chemical reaction of transformation of bauxite into alumina by Hall’s process.
(d) Heating the aluminium ore with concentrated caustic soda.
Answer:
During the extraction of aluminium Chemical reaction of transformation of bauxite into alumina by Hall’s process.

Question 28.
A solution of CuSO₄ in water is ………… in colour.
(a) pink
(b) blue
(c) colourless
(d) green
Answer:
A solution of CuSO₄ in water is blue in colour.

Question 29.
Which of the following process is to be carried out to avoid the formation of greenish layer on brass vessels due to corrosion?
(a) Plating
(b) Anodization
(c) Tinning
(d) Alloying (Practice Activity Sheet – 3)
Answer:
(c) Tinning

State whether the following statements are True or False (If a statement is false, correct it and rewrite it.):

Question 1.
Metals are known as sonar metals.
Answer:
True.

Question 2.
Diamond is the softest natural substance.
Answer:
False. (Diamond is the hardest natural substance.)

Question 3.
Electrolysis method is used to obtain pure metals from impure metals.
Answer:
True.

Question 4.
Iodine and diamond are lustrous substances.
Answer:
True.

Question 5.
Aqua Regia is a mixture of conc. HCl and conc. HNO₃ in the ratio of 1:3.
Answer:
False. (Aqua Regia is a mixture of conc. HCl and conc. HNO₃ in the ratio of 3:1.)

Question 6.
Corrosion of metals can be stopped by detaching the air from metals.
Answer:
True.

Question 7.
Due to corrosion a greenish layer forms on the surface of copper or brass vessel.
Answer:
True.

Question 8.
Ionic compounds are soluble in kerosene.
Answer:
False. (Ionic compounds are soluble in water and insoluble in kerosene.)

Question 9.
Ionic compounds in the solid state conduct electricity.
Answer:
False. (Ionic compounds in the solid state do not conduct electricity.)

Question 10.
Mercury, silver and gold are very reactive metals.
Answer:
False. (Mercury, silver and gold are least reactive metals.)

Question 11.
In electroplating a metal is coated with another metal using electrolysis.
Answer:
True.

Question 12.
In anodising method. the copper or aluminium article is used as anode.
Answer:
True.

Question 13.
Silver plated spoon, gold plated ornaments are the examples of alloying.
Answer:
False. (Silver plated spoon, gold plated ornaments are the examples of electroplating)

Question 14.
silver amalgam is mainly used by dentists.
Answer:
True.

Question 15.
Aluminium oxide is an acidic oxide.
Answer:
False. (Aluminium oxide is an amphoteric oxide.)

Question 16.
copper reacts with moist carbon form copper carbonate.
Answer:
True.

Question 17.
Corrosion is degradation of a reaction with its environment.
Answer:
True.

Find the correlation in the given pair and rewrite the answer:

Question 1.
Brass : Copper and Zinc :: Bronze :………….
Answer:
Brass : Copper and Zinc :: Bronze : Copper and tin

Question 2.
Tinning : Tin :: Galvanizing :…………….
Answer:
Tinning : Tin :: Galvanizing : Zinc

Question 3.
Pressure cooker : Anodizing :: Silver plated spoons :…………..
Answer:
Pressure cooker : Anodizing :: Silver plated spoons : Electro-plating

Question 4.
The sulphides ores are strongly heated in air : Roasting :: The carbonates ores are strongly heated in a limited supply of air :………….
Answer:
The sulphides ores are strongly heated in air : Roasting :: The carbonates ores are strongly heated in a limited supply of air : Calcination.

Question 5.
Sulphide ores : Froth floatation method : Cassiterite ore :………..
Answer:
Sulphide ores : Froth floatation method : Cassiterite ore : Magnetic separation method.

Find the odd one out:

Question 1.
Sodium, Potassium, Silver, Sulphur
Answer:
Sulphur. (All except sulphur, others are metals.)

Question 2.
Boron, Chlorine, Bromine, Fluorine
Answer:
Boron. (All except boron, others are nonmetals.)

Question 3.
Copper, Iron, Mercury, Brass
Answer:
Brass. (All except brass, others are metals.)

Question 4.
Brass, Bronze, Phosphorus, Stainless steel
Answer:
Phosphorus. (All except phosphorus, others are alloys.)

Question 5.
Magnesium chloride, Sodium chloride, Water, Zinc chloride
Answer:
Water. (All except water, others are ionic compounds.)

Question 6.
Tinning, Anodization, Alloying, Froth floatation (March 2019)
Answer:
Froth floatation. (All except froth floatation, others are processes of coating a thin layer of metal on the surface of other metals.)

Match the following:

Question 1.

Answer:
(1) ZnS – Zinc blend
(2) HgS – Cinnabar
(3) Cu₄S – Cuprous sulphide
(4) Al₂O₃.H₂O – Bauxite.

Question 2.

Answer:
(1) Copper and zinc – Brass
(2) Copper and tin – Bronze
(3) Iron, nickel and chromium – Stainless steel
(4) Mercury and zinc – Zinc amalgam.

Question 3.

Answer:
(1) Galvanising – Coating of Zn on iron
(2) Tinning – Coating of tin on copper
(3) Electroplating – Silver plated spoons
(4) Anodizing – Pressure cooker.

Translate the following statements into chemical equations and then balance them:

Question 1.
steam is passed over aluminium.
Answer:

Question 2.
Extraction of copper from its sulphide ore.
Answer:

Question 3.
Thermit reaction.
Answer:

Question 4.
Magnesium reacts with hot water.
Answer:

Question 5.
what happens when aluminium oxide dissolves in aqueous sodium hydroxide?
Answer:

Question 6.
Zinc reacts with sulphuric acid.
Answer:

Question 7.
Iron reacts with sulphuric acid.
Answer:

Name the following:

Question 1.
A metal which forms an amphoteric oxide.
Answer:
Aluminium forms an amphoteric oxide.

Question 2.
An alloy of copper and zinc.
Answer:
An alloy of copper and zinc is termed as brass.

Question 3.
A compound which is added to lower the fusion temperature.
Answer:
Cryolite (AlF₃, 3NaF) and fluorspar (CaF₂) are added to lower the fusion temperature.

Question 4.
A metal which does not react with cold water but reacts with steam.
Answer:
Aluminium does not react With cold water but reacts with Steam.

Question 5.
A common ore of aluminium.
Answer:
Bauxite (Al₂O₃.H₂O) is a common ore of aluminium.

Question 6.
A metal which is in liquid state at ordinary temperature.
Answer:
Mercury is in liquid state at ordinary temperature.

Question 7.
Two metals which are malleable.
Answer:
Iron and aluminium are malleable metals.

Question 8.
Two metals which are ductile.
Answer:
Gold and silver are ductile metals.

Question 9.
Two metals which are good conductors of heat.
Answer:
Silver and copper are good conductors of heat.

Question 10.
Two metals which are good conductors of electricity.
Answer:
Copper and aluminium are good conductors of electricity.

Question 11.
Two metals which are used for making cooking vessels.
Answer:
Copper and aluminium are used in making cooking vessels.

Question 12.
Two metals having low melting points.
Answer:
Sodium and potassium have low melting points.

Question 13.
Two highly reactive metals.
Answer:
Sodium and potassium are highly reactive metals.

Question 14.
A nonmetal which is in liquid state at room temperature.
Answer:
Bromine is in liquid state at room temperature.

Question 15.
Two ionic compounds.
Answer:
Sodium chloride (NaCl) and magnesium chloride (MgCl₂) are ionic compounds.

Question 16.
The process of heating the sulphide ore to a high temperature in the excess of air.
Answer:
In roasting, sulphide ore is heated to a high temperature in the excess of air.

Question 17.
The process of heating the carbonate ore to a high temperature in limited air.
Answer:
In calcination, carbonated ore is heated to a high temperature in limited air.

Question 18.
The compound formed by the reaction between aluminium oxide and sodium hydroxide.
Answer:
Sodium aluminate is formed by the reaction between aluminium oxide and sodium hydroxide.

Question 19.
Two metals which are found in the free state in nature.
Answer:
Gold (Au) and silver (Ag) are found in the free state in nature.

Question 20.
A metal which has the highest melting point.
Answer:
Tungsten has the highest melting point.

Question 21.
Two nonmetals which are lustrous.
Answer:
Iodine and diamond are lustrous in nature.

Answer the following questions in one sentence each:

Question 1.
State the property of the metals due to which they can be drawn into wires.
Answer:
The property of the metal due to which they can be drawn into wires is called ductility.

Question 2.
State the property of the metals due to which they can be beaten into thin sheets.
Answer:
The property of the metals due to which they can be beaten into thin sheets is called malleability.

Question 3.
Which is the hardest substance?
Answer:
Diamond which is a form of carbon is the hardest substance.

Question 4.
What material is used to coat electrical wires?
Answer:
PVC (Polyvinyl chloride) is used to coat electrical wires.

Question 5.
State two metals which can be cut easily with a knife.
Answer:
Sodium and potassium are soft metals and can be cut easily with a knife.

Question 6.
Which of the following metals react with cold water?
Sodium, iron, copper, potassium.
Answer:
Sodium and potassium metals react with cold water.

Question 7.
Which of the following metals do not react with cold water or hot water?
Sodium, potassium, aluminium, iron.
Answer:
Aluminium and iron do not react with cold water or hot water.

Question 8.
State two metals which displace hydrogen from dilute acids and two metals which do not do so.
Answer:
Metals which displace hydrogen from dilute acids are: Magnesium and zinc.
Metals which do not displace hydrogen from dilute acids are: Copper and silver.

Question 9.
Arrange the following metals in the increasing order of their activity:
Copper, Silver, Aluminium, Iron. (Practice Activity Sheet – 1)
Answer:
The arrangement of metals in the increasing order of their activity:
Silver < Copper < Iron < Aluminium

Question 10.
Write the chemical equation for the reaction of hot iron with steam.
Answer:

Question 11.
Complete the following reactions:
(1) Zn_(s) + H₂O_(g) → _____________

Answer:
(1) Zn_(s) + H₂O_(g) → ZnO_(s) + H_2(g)

Question 12.
Complete the following reactions:

Answer:

Question 13.
3MnO₂ + 4Al → 3Mn + 2Al₂O₃ + heat.
Identify the substances undergone oxidation and reduction reactions.
Answer:
MnO₂ is reduced to Mn.
Al is oxidised to Al₂O₃

Question 14.
State the impurities present in the bauxite ore.
Answer:
The main impurities present in the bauxite ore are silica (SiO₂) and iron oxide (Fe₂O₃).

Question 15.
Write the formula of (i) bauxite (ii) cryolite.
Answer:
The formula of bauxite is Al₂O_3·H₂O and that of cryolite is (Na₃AlF₆).

Question 16.
What is galvanization?
Answer:
The process of coating a thin layer of zinc on iron or steel is called galvanization.

Question 17.
Name the reaction in which aluminium is used as a reducing agent.
Answer:
The thermite reaction in which iron oxide is reduced by aluminium. Aluminium is used as a reducing agent in the thermit reaction.

Question 18.
What are the constituents of bronze?
Answer:
Copper and tin are the constituents of bronze.

Question 19.
State the term used to express the purity of gold.
Answer:
The purity of gold is expressed in carat.

Question 20.
What is meant by amalgam?
Answer:
The amalgam is an alloy in which one of the metals is mercury.

Question 21.
What is meant by electroplating?
Answer:
A process in which a less reactive metal is coated on a more reactive metal by electrolysis is called electroplating.

Question 22.
Why are metals called electropositive elements?
Answer:
Metals are reactive. They lose electrons and become positively charged ions. Therefore, metals are called electropositive elements.

Answer the following questions:

Question 1.
Distinguish between the physical properties of metals and nonmetals with respect to the following points:
(1) Physical state (2) Lustre (3) Ductility and malleability (4) Conduction of heat and electricity (5) Hardness (6) Melting and boiling points.
Answer:
(1) Physical state: Under ordinary conditions, metals are generally solids. Exceptions: mercury and gallium are liquids. Under ordinary conditions, nonmetals may be solids or gases. Exception: bromine is in liquid state.

(2) Lustre: Metals usually have a high lustre (called metallic lustre). They can be polished to give a highly reflective surface. With the exceptions of gold and copper, metals usually have silvery grey colour. Nonmetals lack lustre, exceptions: graphite and iodine. Some nonmetals are colourless and others possess a variety of colours.

(3) Ductility and malleability: Metals are ductile and malleable. Nonmetals are not ductile and mallfeable.

(4) Conduction of heat and electricity: Metals are good conductors of heat and electricity. Nonmetals are bad conductors of heat and electricity. Exception: Graphite is a good conductor of electricity.

(5) Hardness: Metals are usually hard, but not brittle, exceptions: sodium, potassium, lead, zinc. Nonmetals are brittle in the solid state, exception: diamond.

(6) Melting and boiling points: The melting and boiling points of metals are high, exceptions: sodium, potassium, mercury, gallium. The melting and boiling points of nonmetals are low, exceptions: carbon, silicon.

Question 2.
Write any three physical properties of nonmetals.
Answer:

1. Nonmetals may be solid or gaseous.
2. Nonmetals lack lustre. They are not ductile and malleable.
3. The melting and boiling points of nonmetals are low.
4. Nonmetals are bad conductors of heat and electricity.

Question 3.
Metals are good conductors of heat. Explain why.
Answer:
(1) The electrons in the outermost shells of atoms of a metal are free to move throughout the metal.
(2) When a metal is heated, these electrons start moving with higher velocity and conduct heat. Hence, metals are good conductors of heat.

Question 4.
Metals are good conductors of electricity. Explain why.
Answer:
(1) The electrons in the outermost shells of atoms of a metal are free to move throughout the metal.
(2) When a potential difference is applied between the ends of a metal wire, the net movement of the electrons in a particular direction, from a point at lower potential to a point at higher potential, constitutes an electric current. Hence, metals are good conductors of electricity.

Question 5.
A metal can be drawn into a wire. Explain why.
Answer:

1. The property due to which a substance can be drawn into a thin wire without cracking or breaking is called ductility.
2. Metals are ductile. Thus, a metal can be drawn into a wire.

Question 6.
A metal can be hammered into a thin sheet. Explain why.
Answer:

1. The property due to which a substance can be hammered (or rolled) into a thin sheet without cracking is called malleability.
2. Metals are malleable. Thus, a metal can be hammered to form a thin sheet.

Question 7.
How do metals react with oxygen?
Answer:
Metals combine with oxygen on heating in air and metal oxides are formed.
Metal + Oxygen → Metal oxide

Question 8.
How does a metal react with water?
Answer:
Sodium and potassium react vigorously with water to evolve hydrogen. Calcium reacts with water slowly and less vigorously to evolve hydrogen and the metal floats on water. Magnesium reacts with hot water to evolve hydrogen. Aluminium, iron and zinc do not react with cold or hot water but they react with steam to evolve their oxides and hydrogen.
2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g) + heat energy
2K_(s) + 2H₂O_(l) → 2KOH_(aq) + H_2(g) + heat energy
2Ca_(s) + 2H₂O_(l) → 2Ca(OH)_2(aq) + H_2(g)
Mg_(s) + 2H₂O_(hot) → Mg(OH)_2(aq) + H_2(g)
2Al_(s) + 3H₂O _steam → Al₂O_3(s) + 3H_2(g)
3Fe_(s) + 4H₂O _steam → Fe₃O₄ + 4H_2(g)
Zn_(s) + H₂O _steam → ZnO_(s) + H_2(g)

Question 9.
(9) How does a metal react with an acid?
Answer:
Reaction of metals with acids: Metals react with dilute hydrochloric acid or dilute sulphuric acid to form metal chloride or metal sulphate and hydrogen gas. The rate of evolution of H2 is maximum in case of magnesium. The reactivity decreases in the order
Mg > Al > Zn > Fe.
Mg_(s) + 2HCl_(aq) → MgCl_2(s) + H_2(g)
2Al_(s) + 6HCl_(aq) → 2AlCl_3(aq) + 3H_2(g)
Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)
Fe_(s) + 2HCl_(aq) → FeCl_2(aq) + H_2(g)
Fe_(s) + H₂SO_4(aq) → FeSO₄ + H_2(g)
Zn_(s) + H₂SO_4(aq) → ZnSO₄ + H_2(g)
Mg_(s) + H₂SO_4(aq) → MgSO_4(aq) + H_2(g)

Question 10.
How does a metal react with nitric acid?
Answer:
Metals react with nitric acid to form nitrate salts. Depending on the concentration of nitric acid, various oxides of nitrogen (NO, NO₂) are formed.

Question 11.
Arrange the following metals in the decreasing order of chemical reactivity:
Cu, Mg, Fe, Ca, Zn, Na.
Answer:
The reactivity of metal decreases in the following order:
Na > Mg > Ca > Zn > Fe > Cu.

Question 12.
What is meant by aqua regia?
Answer:
Aqua regia is a highly corrosive and fuming liquid. It is a freshly prepared mixture of conc. HCl and conc. HNO₃ in the ratio of 3:1. Most of the substances dissolve in it. Aqua regia is a reagent which dissolves gold and platinum.

Question 13.
How does a metal react with salts of other metals?
Answer:
The reaction of metals with solutions of salts of other metals is the displacement reaction. If a metal A displaces other metal B from the solution of its salt, it means that the metal A is more reactive than the metal B.
Metal A + Salt solution of metal B → Salt solution of metal A + metal B

In this reaction Fe has displaced Cu from CuSO₄. It means Fe is more reactive than Cu.

Question 14.
Explain the reactivity series of the metals.
Answer:
The arrangement of metals in decreasing order of their reactivity in the form of a series is called the reactivity or activity series of the metals.
The most reactive metal is placed at the top of the list and least reactive metal is placed at the bottom of the list.

On the basis of reactivity, we can classify metals into the following categories:

1. High reactivity metals
2. Moderately reactive metals
3. Less reactive metals.

1. Extraction of High reactivity metals: The metals which are placed at the top of the reactivity series are very reactive. They are never found in nature as free elements, e.g., sodium, potassium, calcium and aluminium. These metals are obtained by electrolytic reduction.

2. Extraction of Moderately reactive metals: The metals in the middle of reactivity series such as iron, zinc, lead, copper are moderately reactive. These elements are present as sulphides or carbonates in nature. Generally metals are obtained from their oxide as compared to their sulphides and carbonates.

3. Extraction of Less reactive metals: The metals which are placed at the bottom of the reactivity series are least reactive. They occur in free state, e.g. gold, silver and copper. Copper and silver are also found in the combined state as sulphide and oxide ores. These metals are obtained from their ores by just heating the ores in air.

Question 15.
Atomic number of metal “A” is 11, while atomic number of metal “B” is 20. Which of them will be more reactive? Write the chemical reaction of dilute HCl with metal “A”. (Practice Activity Sheet – 2)
Answer:
Metal ‘A’ is more reactive than metal ‘B’.
Atomic number of metal ‘A’ is 11, hence it is Na.
2Na + 2HCl → 2NaCl_(aq) + H_2(g)

Question 16.
How does a metal react with a nonmetal?
Answer:
By oxidation of a metal, cations are formed, on the other hand by reduction of a nonmetal, anions are formed. The ionic compound is formed due to the metal losing electrons while the nonmetal accepts the electrons. The ionic compound of sodium chloride is formed as sodium loses one electron while chlorine accepts one electron.

Similarly Mg and K form ionic compounds MgCl₂ and KCl.

Question 17.
How do nonmetals react with oxygen?
Answer:
Nonmetals combine with oxygen to form acidic oxides. In some cases, neutral oxides are formed.

Question 18.
How do nonmetals react with water?
Answer:
Nonmetals do not react with water, (exception : halogen). Chlorine dissolves in water giving hypochlorous acid.

Question 19.
How do nonmetals react with dilute acids?
Answer:
Nonmetals do not react with dilute acid, (exception: halogen). Chlorine reacts with dil. hydrobromic acid to form bromine and HCl.

Question 20.
How do nonmetals react with hydrogen?
Answer:
Nonmetals react with hydrogen under certain conditions (such as proper temperature, pressure, catalyst, etc.)

Question 21.
What is meant by an ionic compound?
Answer:
The compound formed from two units, namely cation and anion is called an ionic compound.

Question 22.
What is meant by an ionic bond?
Answer:
The cation and anion being oppositely charged, there is an electrostatic force of attraction between them, this force of attraction between cation and anion is called the ionic bond.

Question 23.
State the general properties of ionic compounds.
Answer:

1. Ionic compounds are solids and hard due to strong electrostatic force of attraction between oppositely charged ions.
2. They are generally brittle. When pressure is applied they break into pieces.
3. They have high melting and boiling points, due to intermolecular force of attraction is high in ionic compounds.
4. They are soluble in water and insoluble in solvents such as kerosene and petrol.
5. Ionic compounds cannot conduct electricity when in solid state, they are electrically neutral. They conduct electricity in the molten state and also in an aqueous solution.

Question 24.
Explain the following terms:
1. Concentration of ores
2. Roasting
3. Calcination
4. Refining
Answer:
1. Concentration of ores: The process of separating gangue from the other ores is called concentration of ores.
2. Roasting: The process of heating an ore to a high temperature in excess of air and converting it into its oxide is called roasting.
Examples: ZnS (zinc blend), PbS (Galena)

3. Calcination: The process of heating an ore in a limited supply of air and converting it into its oxide is called calcination.
Example: Zinc carbonate (ZnCO₃)
ZnCO₃ → ZnO + CO₂ ↑
4. Refining: The metal obtained by chemical reduction contains impurities. The process of electrolysis method is used to obtain pure metals from impure metals is known as refining.

Question 25.
State two methods of concentration of ores in which the heavy particles of ores can be separated from the light gangue particles by the gravitational method.
Answer:

1. Wilfley table method
2. Hydraulic separation method are two methods of concentration of ores in which the heavy particles of ores can be separated from the light, gangue particles by the gravitational method.

Question 26.
What are the different methods used for removing gangue from ores?
(OR)
Write the five methods of concentration of ores.
Answer:

1. Wilfley table method
2. Hydraulic separation method
3. Magnetic separation method
4. Froth floatation method
5. Leaching method.

Question 27.
Write short notes on: (1) Wilfley table method (2) Hydraulic separation method (3) Magnetic separation method (4) Froth floatation method (5) Leaching method.
Answer:
(1) Wilfley table method : (Separation based on gravitation) This method of separation uses the Wilfley table, it is made by fixing narrow and thin wooden wedges/blocks on inclined surface with low slope. The table is kept continuously vibrating.

Lumps of the ore is made powdered ore by using ball mill. This powdered ore is poured on the table and a stream of water is simultaneously released from the upper side. This result in the lighter gangue particles getting carried away along with the flowing water, while the heavier particles in which proportion of minerals is more and proportion of gangue particles is less, are blocked by the wooden wedges and is collected through the slits between them.

(2) Hydraulic separation method: The hydraulic separation method is based on the working of a mill. This is a tapering vessel similar to that used in a grinding mill. It opens in a tank like a container that is tapering on the lower side. The tank has an outlet for water on the upper side and a water inlet on the lower side.

Finely ground ore is added to the tank. A fast stream of water is released in the tank from the lower side. The lighter gangue particles flow out along with the water stream from the outlet on the upper side of the tank and are collected separately, simultaneously the heavy particles of the ore are collected at the bottom from the lower side of the tank. This method is based on the law of gravitation, wherein particles of the same size are separated by their weight with the help of water.

(3) Magnetic separation method: Electro-magnetic machine is used in this method. The main parts of this machine are two types of iron rollers and the conveyor belt continuously moving around them. One of the rollers is nonmagnetic while the other is electromagnetic. The conveyor belt moving around the rollers is made up of leather or brass (nonmagnetic). The powdered ore is poured at that end of the conveyor belt which is on the side of the nonmagnetic roller. Two collector vessels are placed below the magnetic roller.

The particles of the nonmagnetic part in the ore are not attracted towards the magnetic roller. Therefore, they are carried out further along the belt and fall in the collector vessel which is away from the magnetic roller. Simultaneously the particles of the magnetic ingredients of the ore stick to the magnetic roller and therefore fall in the collector vessel near the belt.

In this way the magnetic and nonmagnetic particles in the ore are separated because of their magnetic nature. For example, cassiterite is a tin ore. It contains mainly the nonmagnetic ingredient stannic oxide (SnO₂) and the magnetic ingredient ferrous tungstate (FeWO₄). These are separated by the electromagnetic method.

(4) Froth floatation method: The froth floatation method is based on the two opposite properties, hydrophilic and hydrophobic, of the particles. The metal sulphides particles get wet mainly with oil due to their hydrophobic property. The gangue particles get wet with water due to the hydrophilic property.

In this method the finely ground mineral is put into a big tank containing a lot of water. The finely powdered ore and vegetable oil such as pine oil, eucalyptus oil are mixed with water for formation of froth. The pressurised air is blown through the mixture. There is an agitator rotating around its axis in the centre of the floatation tank. The agitator is used as per the requirement. Bubbles are formed due to the blown air.

A foam is formed from oil, water and air bubbles together, due to the agitating. This foam rises to the surface of the water and floats. Hence this method is called froth floatation. Sulphide minerals float with the foam on water as they get and can be removed. The gangue particles are wetted by water, settles down at the bottom. This method is used for concentration of zinc blend (ZnS) and copper pyrite (CuFeS₂).

(5) Leaching: Leaching is the first step in the extraction of the metals like aluminium, gold and silver from their ores. In this method the ore is soaked in a particular solution for long time. The ore dissolves in that solution due to specific chemical reaction. The gangue, however, does not react and therefore does not dissolve in that solution. It can be separated easily.

For example, concentration of bauxite, the aluminium ore, is done by leaching method. Bauxite is soaked in aqueous NaOH or aqueous Na₂CO₃ which dissolves the main ingredient alumina in it. This means that bauxite is leached by sodium hydroxide.

Question 28.
Draw a neat labelled diagram of the arrangement of the equipment used in (1) Wilfley table method (2) Hydraulic separation method.
Answer:
1. Wilfley table method:

2. Hydraulic separation method:

Question 29.
Complete the following flow chart and answer the questions below:

(i) In which method pine oil is used?
(ii) Explain that method of concentration in brief. (Practice Activity Sheet – 2)
Answer:

(i) Pine oil is used in froth floatation method.

(ii) The finely powdered ore and vegetable oil such as pine oil, eucalyptus oil are mixed with water for formation of froth. The pressurised air is blown through the mixture. The agitator is used as per the requirement. Bubbles are formed due to the blown air. A foam is formed from oil, water and air bubbles together, due to the agitating. This foam rises to the surface of the water and floats. Hence this method is called froth floatation. Sulphide minerals float with the foam on water as they get and can be removed. The gangue particles are wetted by water, settles down at the bottom. This method is used for concentration of zinc blend (ZnS) and copper pyrite (CuFeS₂).

Question 30.
A tapping vessel opens in a tank like container that is tapering on the lower side. The tank has an outlet for water on the upper side and a water inlet on the lower side. Finely ground ore is released in the tank. A forceful jet of water is introduced in the tank from lower side and gangue particles and pure ore are separated by this method.
(i) The above description is of which gravitation separation method?
(ii) Draw labelled diagram of this method. (March 2019)
Answer:
(i) Hydraulic separation method.
(ii)

Question 31.
How are sodium, magnesium and potassium obtained from their molten chloride salts?
Answer:
The metals sodium, calcium and magnesium are obtained by electrolysis of their molten chloride salts. In this process metal is deposited on the cathode while chlorine gas is liberated at the anode.

Question 32.
Name the main ore of aluminium.
Answer:
Bauxite (Al₂O₃·H₂O) is the main ore of aluminium.

Question 33.
What is bauxite? What are the main impurities found in this ore?
Answer:
Bauxite (Al₂O₃·H₂O) is hydrated aluminium oxide. It contains 30% to 70% Al₂O₃. The main impurities present in it are iron oxide (Fe₂O₃) and sand (SiO₂).

Question 34.
From which ore is aluminium extracted? What are the stages in its extraction (give only names)?
Answer:
Aluminium is extracted from bauxite (Al₂O₃·nH₂O). Stages id the extraction: (i) Concentration of ore, i.e., conversion of bauxite into alumina, (ii) Electrolytic reduction of alumina.

Question 35.
Describe Bayer’s process for concentration of bauxite.
Answer:
(1) Bayer’s process is used to obtain pure aluminium oxide from bauxite.
(2) Bauxite is then concentrated by chemical separation. Bauxite contains impurities like iron oxide (Fe₂O₃) and silica (SiO₂).
(3) Bauxite ore is powdered and heated with sodium hydroxide under high pressure for 2 to 8 hours at 140 °C in the digester. The aluminium oxide being amphoteric in nature present in bauxite reacts with sodium hydroxide to form water soluble sodium aluminate. This means that bauxite leached by sodium hydroxide. Silica reacts with sodium hydroxide to form soluble sodium silicate. The basic iron oxide (Fe₂O₃) in the gangue remains unaffected. It is separated by filtration.

(4) The filtrate containing sodium aluminate and sodium silicate is stirred with water and then cooling to 50° C. It is hydrolysed to give precipitate of aluminium hydroxide.

(5) Aluminium hydroxide is then filtered, washed with water, dried and then calcinated by heating at 1000 °C to get pure aluminium oxide.

Question 36.
Describe Hall’s process for concentration of bauxite.
Answer:
In Hall’s process the ore is powdered and then it is leached by heating with aqueous sodium carbonate in the digester to form water soluble sodium aluminate. Then the insoluble impurities are filtered opt. The filtrate is warmed and neutralised by passing carbon dioxide gas through it. This result in precipitation of aluminium hydroxide.


The precipitate of Al(OH)₃ obtained in this processes is filtered, washed, dried and then calcinated by heating at 1000 °C to obtain alumina.

Question 37.
Describe the process of preparation of aluminium by the electrolysis of alumina.
(OR)
Draw and label the diagram of electrolysis of alumina and explain the electrolytic reduction of alumina.
Answer:
Electrolytic reduction of alumina:
(1) The electrolytic cell consists of a rectangular steel tank lined from inside with graphite.

(2) The carbon lining (graphite) acts as a cathode. The anode consists of graphite rods suspended in the molten electrolyte.
(3) Alumina has very high melting point ( > 2000 °C). The electrolysis of alumina is carried out at a low temperature by dissolving it in molten cryolite (Na₃AlF₆). The solution of alumina in cryolite and small amount of fluorspar (CaF₂) is added in the mixture to lower its melting point up to 1000 °C.
(4) On passing an electric current, alumina is electrolysed.
(5) Molten aluminium is collected at the cathode, while oxygen gas is evolved at the anode.
The electrode reactions are shown below:
Al₂O₃ → 2Al³⁺ + 3O^2-
Anode reaction: 2O^2- → O_2(g) + 4e^–
Cathode reaction: Al³⁺ + 3e^– → Al
The molten aluminium is heavier than the electrolyte. Therefore, it sinks to the bottom of the electrolyte and is removed from time to time. About 99% pure aluminium is obtained by this process. The oxygen gas liberated reacts with carbon anode and forms carbon dioxide. As the anode gets oxidised during the electrolysis of alumina, it has to be replaced from time to time.

Question 38.
In the extraction of aluminium:
(i) Name the process of concentration of bauxite.
Answer:
The process of concentration of bauxite is known as Bayer’s process.

(ii) Write the cathode reaction in electrolytic reduction of alumina.
Answer:
At the cathode: Al³⁺ + 3e^– → Al.

(iii) Write the function and formula of cryolite in the extraction of aluminium.
Answer:
Cryolite is added to the molten mixture of alumina to reduce the melting point to about 1000 °C.
The formula of cryolite is (Na₃AlF₆) or AlF₃, 3NaF.

(iv) write an equation for the action of heat on aluminium hydroxide.
Answer:

(v) Draw the diagram of extraction of aluminium.
Answer:

(vi) Write the anode reaction in electrolytic reduction of alumina.
Answer:
Al₂O₃ → 2Al³⁺ + 3O^2-
At Anode: 2O^2- → O_2(g) + 4e^–

(vii) Write the cathode reaction in electrolytic reduction of alumina.
Answer:
Al₂O₃ → 2Al³⁺ + 3O^2-
Cathode: Al³⁺ + 3e^– → Al(l)

Question 39.
What happens when aluminium ore is heated with caustic soda? Write the balanced chemical equation for the same.
Answer:
When aluminium ore is heated with caustic soda solution under high pressure for 2 to 8 hours and at 140 °C to 150 °C, aluminium oxide from aluminium ore, being amphoteric in nature, dissolves in caustic soda solution to form sodium aluminate.

Question 40.
How is zinc extracted from its ore zinc sulphide or zinc carbonate?
Answer:
The crude zinc sulphide ore is heated strongly in excess of air. Zinc sulphide is converted into zinc oxide. This process is known as roasting.

(OR)
The crude zinc carbonate ore is heated strongly in limited supply of air. Zinc carbonate is converted into zinc oxide. This process is known as calcination.

The zinc oxide is reduced to zinc by using a reducing agent such as carbon.

Question 41.
How is copper extracted from its sulphide ore?
Answer:
Copper is found as cuprous sulphide (Cu₂S) in nature. When Cu₂S is heated in air, copper is obtained.

Question 42.
How is mercury extracted from cinnabar?
(OR)
Extraction of mercury from its ore cinnabar and write the corresponding chemical reaction.
Answer:
Cinnabar is an ore of mercury. When cinnabar is heated (roasted), it is converted into mercuric oxide (HgO). Mercuric oxide is then reduced to mercury on further heating.

Question 43.
Show the steps involved in the extraction of moderately reactive metals from their sulphide ores.
Answer:
Moderately reactive elements are present as sulphides or carbonates in nature.

Question 44.
In the reactivity series of metals, some metals are misplaced. Rearrange these metals in the decreasing order of their reactivity.

Answer:

Question 45.
Complete the table, if a metal reacts with the reagent then mark ✓ and if not then ✗.

Answer:

Question 46.
Explain the term corrosion with a suitable example.
(OR)
What is corrosion ?
Answer:
The process in which a metal is destroyed gradually by the action of air, moisture or a chemical (like an acid) on its surface is called corrosion.
(OR)
Corrosion is degradation of a material due to reaction with its environment.
The major problem of corrosion occurs with iron, as it is used as a structural material in construction, bridges, shipbuilding.
Iron gets covered by reddish brown flakes when exposed to atmosphere. This is an example of corrosion.

Question 47.
Explain the different methods to prevent corrosion of metals.
Answer:
(1) Corrosion of a metal can be prevented if the contact between metal and air is cut off.
(2) Corrosion of a metal is prevented by coating with something which does not allow moisture and oxygen to react with it.
(3) A layer of oil or paint or grease is applied on the surface of a metal to prevent corrosion. The rusting or corrosion of iron can be prevented by this method.
(4) Corrosion is also prevented by coating a corrosive metal with a noncorrosive metal. Galvanising, tinning, electroplating, anodising and alloying are the different methods in which a metal is coated with a noncorrosive metal to prevent corrosion.

Question 48.
Write three methods of preventing rusting of iron.
Answer:

1. The rusting of iron can be prevented by painting, oiling, greasing or varnishing its surface.
2. Galvanisation is another method of protecting iron from rusting by coating iron with a thin layer of zinc.
3. Corrosion of iron is prevented by coating iron with noncorrosive substance like carbon. This process is termed as alloying.

Question 49.
What is meant by an alloy? Give two examples with chemical composition.
Answer:
The homogeneous mixture formed by mixing a metal with other metals or nonmetals in certain proportion is called an alloy.
Examples:
1. Bronze: Bronze is an alloy formed from 90% copper and 10% tin. Bronze statues stay well in sun and rain.
2. Stainless steel: Stainless steel alloy is made from 74% iron, 18% chromium and 8% carbon. This alloy does not get stained with air or water and does not rust.

Write short notes on the following:

Question 1.
Galvanizing.
Answer:
(1) The process of coating a thin layer of zinc on iron or steel is called galvanization.
(2) In this method corrosion of zinc occurs first because zinc is more electropositive than iron. After a few years zinc layer goes away and the iron layer gets exposed and starts rusting.
(3) In galvanization an iron object is dipped into molten zinc. A thin layer of zinc is formed all over the iron object. Examples: Shiny iron nails, pin, iron pipes.

Question 2.
Tinning.
Answer:
The process of coating a thin layer of tin (molten tin) on copper or brass is called tinning. Cooking vessels made of copper and brass get a greenish coating due to corrosion. The greenish substance is copper carbonate and it is poisonous. If buttermilk or curry is placed in such a vessel it gets spoiled. Therefore, these vessels are coated with tin to prevent corrosion.

Question 3.
Electroplating.
Answer:

The process in which a less reactive metal is coated on a more reactive metal by electrolysis is called electroplating.
Examples: Silver-plated spoon, gold-plated jewellery.
(1) Which process will you study with the help of above material and solutions.
Answer:
With the help of above material and solutions, electroplating process is studied.

(2) Define the process.
Answer:
The process in which less reactive metal is coated on a more reactive metals by electrolysis is called electroplating.

(3) Write the anode and cathode reactions.
Answer:
At anode: Ag → Ag⁺ + e^–
At cathode: Ag⁺ + e^– → Ag

Question 4.
Anodizing.
(OR)
Identify the process shown in the above diagram and explain it in brief. (Practice Activity Sheet – 3)

Answer:
The anodizing technique is an application of electrolysis. In this method copper or aluminium article is used as anode and it is coated with a strong film of their oxides by means of electrolysis. This oxide layer is strong and uniform all over the surface. This thin film protects the metals from corrosion. The protection can be further increased by making the oxide layer thicker during the anodization.
Examples: Kitchen articles such as ; anodized pressure cooker and anodized pan.

Question 5.
Alloying.
Answer:
A homogenous mixture of two or more metals or a metal and a nonmetal in a definite proportion is called an alloy. The physical properties of an alloy are different from those of its constituents. Alloys are corrosion resistant. Alloy decreases the intensity of corrosion of metals.

Examples: Brass is made from copper and zinc, 90 % Copper and 10 % tin are used to make an alloy called bronze, Stainless steel is made from 74% iron, 8 % carbon and 18 % chromium.

Distinguish between: (Two points of distinction)

Question 1.
Metals and Nonmetals.
Answer:
Metals:

1. Metals have a lustre.
2. They are generally good conductors of heat and electricity.
3. They are generally solids at room temperature.
   Exception: Mercury and gallium are liquids.
4. Metals form basic oxides.

Nonmetals:

1. Nonmetals have no lustre.
   Exception: Iodine and Diamond.
2. They are bad conductors of heat and electricity.
   Exception: Graphite.
3. They are generally gases and solids at room temperature.
   Exception: Bromine is a liquid.
4. Nonmetals form acidic or neutral oxides.

Question 2.
Roasting and Calcination.
Answer:
Roasting:

1. In this process, the ore is heated strongly in the presence of air.
2. In this process, sulphide ore is converted into metal oxide.
3. During this process SO₂ is given out.

Calcination:

1. In this process, the ore is heated strongly in the limited supply of air.
2. In this process, carbonate ore is converted into metal oxide.
3. During this process CO₂ is given out.

Give scientific reasons for the following:

Question 1.
Calcium floats on water during the reaction with water.
Answer:

1. Calcium reacts with water less vigorously hence the heat evolved is not sufficient for hydrogen to catch fire.
2. Instead, calcium floats on water because the bubbles of hydrogen gas formed stick to the surface of the metal.

Question 2.
Common salt has high melting and boiling points.
Answer:

• Common salt is an ionic compound. Common salt is solid and hard due to strong electrostatic attraction between oppositely charged Na+ and Cl- ions.
• A large amount of energy is required to break the strong intermolecular attraction (strong ionic bond). Hence, common salt has high melting and boiling points.

Question 3.
Metals are good conductors, while non-metals are poor conductors of electricity.
Answer:

• The electrons in the outermost orbit of atoms of a metal are free to move throughout the metal.
• When a potential difference is applied between the ends of a metal wire, the movement of the electrons constitutes an electric current. Hence, metals are good conductors of electricity.
• Nonmetals involve covalent bonding and do not have free electrons like metals. Hence, nonmetals are poor conductors of electricity.

Question 4.
Sodium is more reactive than aluminium.
Answer:

• If the number of electrons in the outermost orbit of an atom of a metal is less, the metal is more reactive.
• Sodium has electronic configuration (2, 8, 1) and aluminium has electronic configuration (2, 8, 3). The number of electrons in the outermost orbit of sodium and aluminium atoms are 1 and 3, respectively. Hence, sodium is more reactive than aluminium.

Question 5.
When zinc granules are added to copper sulphate solution, the blue coloured solution turns colourless.
Answer:

• Zinc is more reactive than copper.
• When zinc granules are added to copper sulphate solution, they displace copper from the copper sulphate solution to form zinc sulphate solution. As zinc sulphate is colourless, the blue coloured solution of copper sulphate disappears.

Question 6.
When an iron nail is dipped into a copper solution, a shiny coat of copper is formed on the nail.
Answer:

• Iron is more reactive than copper.
• When an iron nail is dipped into copper sulphate solution, iron displaces copper from the copper sulphate solution. The copper so liberated deposits on the iron nail. As a result, a shiny coat of copper is formed on the nail.

Question 7.
Cryolite (Na3AlF6) and fluorspar (CaF₂) are added to the electrolytic mixture containing pure alumina.
Answer:
(1) Alumina has very high melting point ( > 2000 °C). Cryolite (Na₃AlF₆) and fluorspar (CaF₂) lower the fusion temperature of the mixture containing alumina from 2000 ° C to 1000 ° C, thereby saving electrical energy.
(2) They increase the conductivity and the mobility of the fused mixture. Hence, cryolite and fluorspar are added to the electrolytic mixture containing pure alumina.

Question 8.
Air is bubbled through the mixture in Froth floatation process.
Answer:
(1) In the froth floatation process, in a tank water, ore and an oil are mixed. When air is bubbled through the mixture the oil forms froth.
(2) The mineral particles are wetted by the oil and float on the surface.
(3) The gangue particles are wetted by water and settle down. Hence, the ore can be concentrated. Hence, air is bubbled through the mixture in froth floatation process.

Question 9.
Silver amalgam is used for filling dental cavities.
Answer:
(1) Silver is a soft metal and wears off on constant usage particularly due to abrasion. Silver amalgam is an alloy of silver with mercury.
(2) It is a hard substance. It is nontoxic. Besides these properties it is a lustrous shining substance. It melts at a comparatively low temperature and can therefore conveniently fill in the cavities. Hence, silver amalgam is used for filling dental cavities.

Explain the following reactions with the help of balanced equations:

Question 1.
Out of sodium and sulphur which is a metal? Explain its reaction with oxygen. (March 2019)
Answer:
Sodium is a metal. Sodium reacts with oxygen in air at room temperature to form sodium oxide.

Question 2.
Magnesium burns in air.
Answer:
When magnesium bums in air, it combines with oxygen, emitting intense light and heat to form magnesium oxide.

Question 3.
Copper reacts with air.
Answer:
Copper tarnishes in moist air and forms black coloured oxide when strongly heated.

Question 4.
Sodium reacts with water.
Answer:
When sodium reacts with water, it evolves hydrogen which immediately catches fire producing a lot of heat.

Question 5.
Calcium reacts with water.
Answer:
Calcium reacts with water less vigorously to form hydrogen gas and calcium hydroxide. In this reaction, the heat evolved is not sufficient for hydrogen to catch fire.

Question 6.
Steam is passed over aluminium.
Answer:
When steam is passed over aluminium, hydrogen gas and aluminium oxide are formed.

Question 7.
Steam is passed over iron.
Answer:
When steam is passed over iron, iron (III) oxide and hydrogen gas are formed.

Question 8.
Magnesium reacts with dil. hydrochloric acid.
Answer:
When magnesium reacts with dil. hydrochloric acid, magnesium chloride is formed and hydrogen gas is evolved.

Question 9.
Aluminium is treated with dil. hydrochloric acid.
Answer:
When aluminium is treated with dil. hydrochloric acid, aluminium chloride and hydrogen gas are formed.