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Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 3 Ionic Equilibria Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 3 Ionic Equilibria

Question 1.
What are electrolytes?
Answer:
Electrolytes: The substances which in their aqueous solutions (or in any polar solvents) dissociate or ionize forming positively charged ions (cations) and negatively charged ions (anions) are called electrolytes. For example, NaCl, HCl, etc.

Question 2.
What is an ionic equilibrium?
Answer:
Ionic equilibrium: The equilibrium between ions and unionized molecules of an electrolyte in solution is called an ionic equilibrium.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+}\)

Question 3.
What are the types of electrolytes?
Answer:
There are two types of electrolytes as follows :
(a) Strong electrolyte The electrolytes which ionise completely or almost completely are called strong electrolytes. For example, NaCl, HCl, H₂SO₄, etc.
(b) Weak electrolytes: The electrolytes which dissociate to a less extent are called weak electrolytes. For example, CH₃COOH, NH₄OH, etc.

Question 4.
Define degree of dissociation.
Answer:
Degree of dissocsavIt is defined as a fraction of total number of moles of an electrolyte that dissociate into its ions at equilibrium.
It is denoted by a and represented by,

OR α = \(\frac{\text { Per cent dissociation }}{100}\)
∴ Per cent dissociation = α × 100

Question 5.
Define acid and base. Give examples.
Answer:
Acid : A hydrogen containing substance which gives H⁺ ions in aqueous solution is called an acid. For example, HCl, CH₃COOH, etc.
\(\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)

Base : A substance that contains OH group and produces hydroxide ions (OH^–) in aqueous solution is called a base.
\(\mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)

Question 6.
What are the limitations of Arrhenius acid-base theory ?
Answer:
Limitations of Arrhenius theory :

1. This theory is applicable only for aqueous solutions and not for non-aqueous solutions.
2. It fails to explain the acidic nature of non-hydrogen compounds like BF₃, AlCl₂, FeCl₃, etc.
3. It fails to explain the basic nature of non-hydroxy compounds like NH₃, amines, Na₂CO₃, KCN, aniline, etc. in their aqueous solutions.
4. It does not explain role of solvent or existence of H₃O⁺ in an aqueous solution of an acid.

Question 7.
Explain neutralisation reaction according to Arrhenius theory.
Answer:
Neutralisation reaction : According to Arrhenius theory neutralisation is a reaction between an acid and a base in their aqueous solutions produciijg salt and unionised water.
\(\mathrm{HCl}_{(\mathrm{aq})}+\mathrm{NaOH}_{(\mathrm{aq})} \rightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
Since strong acid, strong base and salt dissociate completely, the above reaction is represented as,

Hence, according to Arrhenius theory neutralisation reaction is defined as a reaction between H⁺ ions and OH^– ions forming unionised water molecules.

Question 8.
Explain Bronsted-Lowry theory of acids and bases.
Answer:
Acid : According to Bronsted-Lowry theory acid is a substance that donates a proton (H⁺) to another substance.
Base : According to this theory base is a substance that accepts a proton (H⁺) from another substance. For example,

Since HCl donates a proton it is an acid while NH₃ accepts a proton it is a base.

Question 9.
For each of the following reactions, identify the Lowry-Bronsted conjugate acid-base pairs :

Answer:

In this Acid-1 (CH₃COOH) and Base-1 (CH₃COO^–) is one acid base conjugate pair while Base-2 (NH₃) and Acid-2 \(\mathrm{NH}_{4}^{+}\) is another conjugate pair.

Question 10.
Mention a conjugate acid and a conjugate base for each of the following :
(a) H₂O (b) \(\mathrm{HSO}_{4}^{-}\) (C) Br^– (d) H₂CO₃ (e) \(\mathbf{H}_{2} \mathbf{P O}_{4}^{-}\) (f) \(\mathbf{N H}_{4}^{+}\)
Answer:

Question 11.
Identify conjugate acid-base pairs in the following :


Answer:

Question 12.
Define acids and bases on the basis of Lewis concept. Give examples.
Answer:
Lewis concept of an acid and a base is based on the electronic theory.
Acid : It is defined as any species (molecule or ion) that can accept a pair of electrons. E.g. BF₃, AlCl₃ and all electron deficient species like cations (K⁺, Ag⁺) and molecules having incomplete octet, like BeF₂, BF₃.

Base : It is defined as any species (molecule or ion) that can donate a pair of electrons. E.g. NH₃, C₂H₅NH₂ and all electron rich species like anions (Cl^–, OH^–) and all molecules with lone pair of electrons.

Question 13.
Explain : (A) BF₃ is a Lewis acid, (B) NH₃ is a Lewis base.
Answer:
(A) According to Lewis theory, an acid is a substance which can accept a pair of electrons.
In BF₃ molecule, the octet of B is incomplete, hence it needs two electrons or a pair of electrons to complete its octet. Hence BF₃ acts as a Lewis acid.

(B) According to Lewis theory, a base is a substance which can donate a pair of electrons.
In NH₃ molecule, nitrogen atom has one lone pair of elctrons to donate.

The reaction between BF3 and NH3 can be represented as,

Question 14.
Classify the following into Lewis acids and bases :
CN^–, Cl^–, S^2-, Cu⁺⁺, H₂O, OH^–, BF₃, Ag⁺.
Answer:
(1) Lewis acid : Cu⁺⁺, BF₃, Ag⁺
(2) Lewis bases : CN^–, Cl^–, S^2-, OH^–.

Question 15.
Explain amphoteric nature of water.
Answer:
(1) Since water acts as an acid as well as a base, it is amphoteric in nature.
(2) H₂O has a tendency to donate a proton forming OH^– as well as has a tendency to accept a proton forming H₃O⁺.


Therefore H₂O is amphoteric in nature.

Question 16.
How are acids and bases classified on the basis of extent of their dissociation?
Answer:
On the basis of extent of dissociation acids and bases are classified as follows :
(1) Strong acids and strong bases : The acids and bases which dissociate to a greater extent or almost completely are called strong acids and strong bases.
\(\mathrm{HCl}_{\text {(aq) }} \longrightarrow \mathrm{H}_{\text {(aq) }}^{+}+\mathrm{Cl}_{\text {(aq) }}^{-}\)
\(\mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)

(2) Weak acids and weak bases : The acids and bases which dissociate partially are called weak acids and weak bases. There exists an equilibrium between undissociated molecules and ions in solution.

Question 17.
Give examples of (a) strong acids and strong bases, (b) weak acids and bases.
Answer:
(a) Strong acids : HCl, H₂SO₄
Strong bases : NaOH, KOH
(b) Weak acids : HCOOH, CH₃COOH
Weak bases : NH₄OH, C₂H₅NH₂

Question 18.
Define and explain dissociation constant of a weak acid.
Answer:
Dissociation constant of a weak acid : It is defined as the equilibrium constant for dissociation equilibrium of a weak acid and denoted by K_a.
Explanation : Consider an aqueous solution of a weak acid HA.
\(\mathrm{HA}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}\)
The equilibrium constant called dissociation constant Ka is represented as,
K_a = \(\frac{\left[\mathrm{H}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)

Question 19.
Derive the expression of Ostwald’s dilution law in case of a weak acid (HA).
OR
Derive the relationship between degree of dissociation and dissociation constant of a weak acid.
Answer:
Expression of Ostwald’s dilution law in case of a weak acid : Consider the dissociation of a weak acid HA. Let V dm³ of a solution contain one mole of weak acid HA. Then the concentration of a solution is, C = \(\frac{1}{V}\) mol dm^3-3. Let α be the degree of dissociation of HA.

Applying the law of mass action to this dissociation equilibrium, we have,

As the acid is weak, a is very small as compared to unity,
∴ (1 – α) ≈ 1.

This is an expression for Ostwald’s dilution law. This shows that the degree of dissociation of a weak acid is directly proportional to the volume of solution containing one mole of acid or inversely proportional to square root of its concentration.

Question 20.
Derive the relationship between degree of dissociation and dissociation constant of a weak base.
OR
Derive the expression of Ostwald’s dilution law in case of a weak base.
Answer:
Expression of Ostwald’s dilution law in case of a weak base : Consider the dissociation of a weak base BOH. Let V dm³ of a solution contain one mole of weak base BOH. Then the concentration of a solution is, C = \(\frac{1}{V}\) mol dm^-3. Let α be the degree of dissociation of BOH.

Consentration at equilibrium (mol dm^-3) \(\frac{(1-\alpha)}{V} \quad \frac{\alpha}{V} \quad \frac{\alpha}{V}\)
Applying the law of mass action to this dissociation equilibrium, we have,

This is an expression of Ostwald’s dilution law. Thus, the degree of dissociation of a weak base is directly proportional to the square root of the volume of the solution containing one mole of a base or inversely proportional to the square root of its concentration.

Solved Examples 3.4

Question 21.
Solve the following:

(1) At 298 K, 0.01 M formic acid solution is 1.2% dissociated. Calculate the dissociation constant of formic acid.
Solution :
Given : Concentration of HCOOH = C = 0.01 M
Percent dissociation = 1.2
Dissociation constant = K_a = ?
∴ Degree of dissociation = \(\frac{\text { Percent dissociation }}{100}\)
α = \(\frac{1.2}{100}\)
= 1.2 × 10^-2
K_a = Cα²
= 0.01 × (1.2 × 10^-2)²
= 1.44 × 10^-6
Ans. Dissociation constant = K_a = 1.44 × 10^-6

(2) The degree of dissociation of ammonium hydroxide is 0.0232 in 0.5 M solution. What will be the dissociation constant of ammonium hydroxide ?
Solution :
Given : Degree of dissociation = α = 0.0232
Concentration of NH₄OH = C = 0.5 M
Dissociation constant = K_b = ?
K_b = Cα²
= 0.5 × (0.0232)²
= 2.692 × 10^-4
Ans. Dissociation constant of NH₄OH
= K_b = 2.692 × 10^-4.

(3) Calculate the hydrogen ion concentration in 0.1 M acetic acid solution when the acetic acid is 2% dissociated in the solution.
Solution : The dissociation of acetic acid is represented below :
CH₃COOH ⇌ CH₃ – COO^– + H⁺
Given : Dissociation = 2%, C = 0.1 M
[H₃O⁺] = ?
The concentration of hydrogen ion, [H⁺], is given by the following formula :
[H₃O⁺] = αC
α = Degree of dissociation = \(\frac{\text { Per cent dissociation }}{100}\)
= \(\frac{2}{100}\) = 0.02
[H₃O⁺] = Hydrogen ion concentration = ?
C = Molar concentration of acetic acid
= 0.1 M = 0.1 mol dm^-3
∴ [H₃O⁺] = C × α = 0.1 × 0.02
= 0.002 = 2.0 × 10^-3 mol dm^-3
Ans. Hydrogen ion concentration = 2.0 × 10^-3 mol dm^-3.

(4) Calculate the percentage dissociation of 0.01 M NH₄OH solution. The K_b for NH₄OH is 1.75 × 10^-5.
Solution :
Given : Concentration of NH₄OH = C = 0.01 M
Dissociation constant of NH₄OH
= K_b= 1.75 × 10^-5
Percentage dissociation = ?
K_b = Cα²

∴ Per cent dissociation = α × 100
= 4.183 × 10^-2 × 100
= 4.183
Ans. Dissociation of NH₄OH = 4.183%

(5) 0.01 mole of a weak base is dissolved in 8 dm³ of water. The dissociation constant of the base is 4.0 × 10^-10. Calculate the degree of dissociation of the base in the solution.
Solution :
Given : V = 8 dm³; n = 0.01 mole; K_b = 4 × 10^-10
α = ?
The degree of dissociation and dissociation constant of a weak base are related to each other by the following formula :
K_b = α²C OR α = \(\sqrt{\frac{K_{b}}{C}}\)
K_b = Dissociation constant of the base = 4.0 × 10^-10
α = Degree of dissociation of the base = ?
C = Molar concentration of the base
= 0.01 mole in 8 dm³

Ans. Degree of dissociation of the base = 5.656 × 10^-4

(6) The dissociation constant of benzoic acid (C₆H₅COOH) is 6.6 × 10^-5. Calculate the hydrogen ion concentration of a solution containing 1.22 g of benzoic acid in 2000 mL of water.
Solution :
Given : K_a = 6.6 × 10^-5; V = 2000 mL;
W = 1.22 g; [H⁺] = ?
Molar mass of benzoic acid (C₆H₅COOH) = 122
The concentration of the solution is 1.22 g benzoic acid in 2000 ml (2 dm³) of solution.
1.22 g = \(\frac{1.22}{122}\) = 0.01 mol
∴ Molar concentration of benzoic acid = \(\frac{0.01}{2}\)
= 0.005 mol dm^-3
The dissociation constant and degree of dissociation of a weak acid are,
K_a = α²C OR α = \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
α = Degree of dissociation of benzoic acid = ?
C = 0.005 mol dm^-3

Since benzoic acid is a monobasic acid, [H⁺] = aC
∴ [H⁺] = 0.1149 × 0.005
= 5.745 × 10^-4 mol dm^-3
Ans. Hydrogen ion concentration
= 5.745 × 10^-4 mol dm^-3

(7) The degree of dissociation of acetic acid in its 0.1 M solution is 0.0132 at 25 °C. Calculate the degree of dissociation in its 0.01 M solution.
Solution :
Given : C = 0.1 M, α = 0.0132, C’ = 0.01 M,
α = 0.0132, α’= ?
K_a = α²C
C = Molar concentration of acetic acid = 0.1 M
α = Degree of dissociation in 0.1 M solution = 0.0132
∴ α = \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
If the concentration is C’, α’ = \(\sqrt{\frac{K_{\mathrm{a}}}{C^{\prime}}}\)

Ans. Degree of dissociation in 0.01 M solution = 4.175 × 10^-2

Question 22.
Explain autoionisation of water. Derive a relation for ionic product of water.
Answer:
Pure water ionises to a very less extent. The ionisation equilibrium is represented as follows,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The equilibrium constant K for the above equilibrium is represented as,
K = \(\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}}\)
∴ K × [H₂O]² = [H₃O⁺] × [OH^–]
Since K and active mass of pure water [H₂O] are constant we can write,
K × [H₂O] = K_w,
∴ K_w= [H₃O⁺] × [OH^–]
where K_w is called ionic product of water. At 25 °C,
K_w= 1 × 10^-14.

Question 23.
Define ionic product of water.
Answer:
Ionic product of water : It is defined as the product of molar concentrations of hydronium ions (or hydrogen ions) and hydroxyl ions at equilibrium in pure water at constant temperature.
It is represented as,
K_w = [H₃O⁺] × [OH^–]
At 25 °C, K_w= 1 × 10^-14.

Question 24.
Define the following :
(i) pH (2) pOH. (2 marks)
Ans.
(1) pH : The negative logarithm, to the base 10, of the molar concentration of hydrogen ions, H⁺ is known as the pH of a solution.
PH = -log₁₀ [H⁺]
(2) pOH : The negative logarithm, to the base 10, of the molar concentration of hydroxyl ions, OH^– is known as the pOH of a solution.
pOH = -log₁₀ [OH^–]

Question 25.
What are approximate concentrations of H₃O⁺ and OH^– in, (a) pure water or neutral solution, (b) acidic solution and (c) basic solution ? Also mention pH values.
Answer:

Question 26.
Write a note on pH scale.
Answer:
Most of the chemical reactions and industrial processes are carried out in aqueous solutions, hence there is a need to know concentration of H⁺ and OH^– ions in the solution.
Sorensen developed a convenient scale to represent the acidic, basic or neutral nature of the solution.
The pH scale is used to express the concentration of H⁺ and OH^– along with pH and pOH of the solution.
According to Sorensen,
pH = -log₁₀ [H+], pOH = -log₁₀ [OH^–]
pH + pOH = 14.

Acids, basic and neutral solutions.

Question 27.
Solve the following :

(1) At 50 °C the value of ionic product of water is 5.5 × 10^-14. What are the concentrations of [H₃O⁺] and OH^– in a neutral solution at 50°C temperature ?
Solution :
Given : at 50 °C
Ionic product of water = K_w = 5.5 × 10^-14
[H₃O⁺] = ? OH^– = ?
Water at any temperature will be neutral.
Hence, [H₃O⁺] = [OH^–] = x mol dm^-3
[H₃O⁺] × [OH^–] = K_w
x × x = 5.5 × 10^-14
∴ x = 2.345 × 10^-7
∴ [H⁺] = [OH^–] = 2.345 × 10^-7 M
Ans. Concentrations : [H₃O⁺] = [OH^–]
= 2.345 × 10^-7 M

(2) The concentration of H+ ion in lemon juice is 2.5 × 10^-3 M. Calculate the OH^– ion concentration and classify the solution as acidic, basic or neutral.
Solution :
Given : [H₃O⁺] = 2.5 × 10^-3 M, K_w = 1 × 10^-4
[OH^–] = ?
By ionic product of water,
[H₃O⁺] × [OH^–] = K_w
∴ [OH^–] = \(\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\)
= \(\frac{1 \times 10^{-14}}{2.5 \times 10^{-3}}\)
Ans. Concentration of OH^–
= [OH^–] = 4 × 10^-12 M
Hence the solution of lemon juice is acidic.

(3) Calculate pH and pOH of 0.02 M HCl solution.
Solution :
Given : C = 6.02 M HCl; pH = ? pOH = ?
\(\begin{aligned}
\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow & \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \\
& 0.02 \mathrm{M}
\end{aligned}\)
[H⁺] = [H₃O⁺] = 0.02 M
PH= -log₁₀ [H₃O⁺]
= -log₁₀ 0.02
= -(\(\overline{2} .3010\))
= 2 – 0.3010 = 1.699
pH + pOH = 14
∴ pOH = 14 – pH
= 14 – 1.699
= 12.3010
Ans. pH = 1.6990; pOH = 12.3010.

(4) The pH of a solution is 6.06. Calculate its [H₃O⁺] ion concentration.
Solution :
Given : pH = 6.06, [H₃O⁺] = ?
PH = -log₁₀ [H₃O⁺]
∴ log₁₀ [H₃O⁺] = -pH
∴ [H₃O⁺] = Antilog – pH
= Antilog – 6.06
= Antilog \(\overline{7} .94\)
= 8.714 × 10^-7 M
Ans. [H₃O⁺] = 8.714 × 10^-7 M.

(5) Calculate number H⁺ ions present in 1 mL of 0.01 M H₂SO₄ solution.
Solution :
Given : C = 0.01 M H₂SO₄; Y = 1 mL
Number of H⁺ ions = ?
\(\begin{aligned}
\mathrm{H}_{2} \mathrm{SO}_{4(\mathrm{aq})} \longrightarrow & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \\
& 0.01 \times 2 \mathrm{M}
\end{aligned}\)
∵ 1000 mL solution contains 0.02 mol H⁺
∴ 1 mL solution contains \(\frac{0.02}{1000}\) mol
= 2 × 10^-5 mol H⁺
∴ Number of H⁺ ions = 2 × 10^-5 × 6.022 × 10²³
= 1.204 × 10¹⁹
Ans. Number of H⁺ ions = 1.204 × 10¹⁹

(6) The pH of a 0.1 M monoacidic base is 11.11. What is the percent dissociation of base ?
Solution :
Given : pH = 11.11; per cent Dissociation of base = ?
c = 0.1 M
\(\begin{gathered}
\mathrm{BOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \\
c(l-\alpha) \quad c \alpha \quad c \alpha
\end{gathered}\)
pH + pOH = 14
∴ pOH = 14 – pH= 14 – 11.11 = 2.89
POH = -log₁₀ [OH^–]
∴ [OH^–] = Antilog – pOH
= Antilog – 2.89
= Antilog \(\overline{3} .11\)
= 1.29 × 10^-3 M
∵ [OH^–] = cα
∴ α = \(\frac{\left[\mathrm{OH}^{-}\right]}{c}=\frac{1.29 \times 10^{-3}}{0.1}\) = 1.29 × 10^-2
∴ Per cent dissociation = α × 100
= 1.29 × 10^-2 × 100 = 1.29
Ans. Per cent dissociation = 1.29.

(7) Calculate the pH of dedmolar solution of sulphuric add aqueous solution. Assuming complete Ionization of sulphuric add.
Solution:
Given : Concentration of H₂SO₄ = decimolar
= 0.1 M
H₂SO₄ → 2H⁺ + \(\mathrm{SO}_{4}^{-2}\)
∴ [H⁺] = 2 × 0.1
= 0.2 M
PH = -log₁₀ [H⁺]
= -log₁₀ 0.2 M
= \(-[\overline{1} .3010]\)
= 1 – 0.3010 = 0.6990
Ans. pH of H₂SO₄ solution = 0.6990.

(8) The pH of a 0.2 M solution of ammonia is 10.78. Calculate (i) OH^– ions concentration (ii) the degree of dissociation (iii) the dissociation constant.
Solution :
Given: pH = 10.78, C = 0.02 M, [OH^–] = ? K_b = ?
As NH₃ is a base,
pOH = 14 – pH
pH = 10.78
∴ pOH = 14 – 10.78 = 3.22
pOH = -log₁₀ [OH^–]
∴ 3.22 = -log₁₀ [OH^–]
∴ -3.22 = log₁₀ [OH^–]
[OH^–] = antilog (-3.22)
∴ [OH^–] = antilog \((\overline{4} .78)\)
= 6.026 × 10^-4 M
As NH₄OH is a monoacidic base, [OH^–]

(9) NH₄OH is 4.3% ionised at 298 K in 0.01 M solution. Calculate the ionization constant and pH of NH₄OH.
Solution :
Given : Per cent dissociation = 4.3,
C = 0.01 M, K_b = ?, pH = ?
The degree of dissociation and dissociation constant of NH₄OH are related to each other by the formula :
K_b = α²C
K_b = Dissociation constant of NH₄OH = ?
α = Degree of dissociation of NH₄OH = 4.3%
= 4.3 × 10^-2
C = Molar concentration of NH₄OH = 0.01 M
∴ K_b = (4.3 × 10^-2)² × 0.01
= 18.49 × 10^-4 × 10^-2
∴ K_b = 1.849 × 10^-5
Since NH₄OH is a monoacidic base,
[OH^–] = αC
= 4.3 × 10^-2 × 0.01
= 4.3 × 10^-4 mol dm^-3
pOH = -log₁₀ [OH-]
= -log₁₀ 4.3 × 10^-4
= -[0.6335 – 4] = 3.3665
pH + pOH = 14
pH = 14 – 3.3665 = 10.6335
Ans. K_b = 1.849 × 10^-5, pH = 10.6335

Question 28.
Define hydrolysis.
Answer:
Hydrolysis : A reaction in which the cations or anions or both the ions of a salt react with water to produce acidity or basicity or sometimes neutrality is called hydrolysis.

Question 29.
What are the types of the salts? Give examples.
Answer:
A salt is formed by the reaction between equivalent amounts of an acid and a base. According to the nature of an acid and a base, there are four types of the salts as follows :
(1) Salt of a strong acid and a strong base :

(2 ) Salt of a weak acid and a strong base :

(3) Salt of a strong acid and a weak base :

(4) Salt of a weak acid and a weak base : CH₃COONH₄ :

Question 30.
A salt of strong acid and strong base does not undergo hydrolysis. Explain.
OR
An aqueous solution of sodium chloride is neutral. Explain.
Answer:
(1) Sodium chloride is a salt of strong acid HCl and strong base NaOH.
(2) In water, it reacts forming HCl and NaOH.
(3) As both are strong, they dissociate almost completely to liberate H⁺ and OH^– ions, respectively.
(4) H⁺ and OH^– ions combine together to form weakly dissociating H₂O. As there are no free H⁺ ions and OH^– ions, the solution is neutral and the salt does not undergo hydrolysis.
NaCl + H₂O ⇌ NaOH + HCl
Ionic equation :
Na⁺ + Cl^– +H₂O ⇌ Na⁺ + OH^– + H⁺ + Cl^–
H₂O ⇌ H⁺ + OH^–
Since the solution contains equal number of H⁺ and OH^– ions, it is neutral.
Hence the salt of strong acid and strong base does not undergo hydrolysis.

Question 31.
Explain the hydrolysis of the salt of strong acid and weak base.
OR
A solution of CuSO₄ reacts acidic. Explain.
Answer:
(1) Consider a salt of strong acid and weak base, like CuSO₄ obtained from strong acid H₂SO₄ and weak base Cu(OH)₂.
(2) When it is dissolved in water it undergoes hydrolysis as follows :

(3) Since [H₃O⁺] > [OH^–], the solution is acidic in nature.

Question 32.
Explain the hydrolysis of a salt of weak acid and strong base.
OR
A solution of sodium acetate, CH₃COONa reacts basic explain.
Answer:
(1) Consider a salt of weak acid and strong base like CH₃COONa. In aqueous solution it undergoes hydrolysis as follows :

(2) Since the base NaOH is strong, it dissociates completely while acid CH₃COOH being weak dissociates partially.
(3) Hence [OH^–] > [H₃O⁺] in the solution and the solution reacts basic.

Question 33.
Explain the hydrolysis of the salt of weak acid and weak base.
Answer:
(1) Consider a salt BA of weak acid (HA) and weak base (BOH).
(2) In aqueous solution it undergoes hydrolysis as follows :

(3) The nature of the solution will depend upon relative strength of weak acid and weak base, hence will depend upon their dissociation constants K_a and K_b.

(i) A salt of weak acid and weak base for which K_a > K_b :
Consider hydrolysis of NH₄F.

Since K_a (7.2 × 10^-4) for HF is greater than K_b (1.8 × 10^-5) for NH₄OH, the acid dissociates partially more than the base, hence, [H₃O⁺] > [OH^–] and the solution reacts acidic after hydrolysis.

(ii) A salt of weak acid and weak base for which K_a < K_b :
Consider hydrolysis of NH₄CN.

Since K_a (4 × 10^-10) for HF is less than K_b (1.8 × 10^-5) for NH₄OH, the base dissociates more than acid and hence [H₃O⁺] < [OH^–] and the solution reacts basic after hydrolysis.

(iii) A salt of weak acid and weak base for which K_a = K_b:
Consider hydrolysis of CH₃COONH₄.

Since K_a = K_b, the weak acid CH₃COOH and weak base NH₄OH dissociate to the same extent, hence, [H₃O⁺] = [OH^–] and the solution reacts neutral after hydrolysis.

Question 34.
Define buffer solution.
OR
What is buffer solution?
Answer:
Buffer solution : It is defined as a solution which resists the change in pH even after the addition of a small amount of a strong acid or a strong base or on dilution or on addition of water.

Question 35.
What are the types of buffer solutions ?
Answer:
These are two types of buffer solutions :
(A) Acidic buffer solution : it is a solution containing a weak acid e.g. (CH₃COOH) and its salt of a strong base. e.g. (CH₃COONa).
pH of an acidic buffer is given by following Henderson Hasselbalch equation,
pH = \(\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}\)
where pK_a = -log₁₀ K_a
and K_a is the dissociation constant of weak acid.

(B) Basic buffer solutions : It is a solution containing a weak base (e.g. NH₄OH) and its salt of strong acid, (e.g. NH₄Cl).
pOH of a basic buffer is given by Henderson Hassebalch equation,
pOH = \(\mathrm{p} K_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}\)
where pK_b = -log10 K_b
and K_b is the dissociation constant of a weak base.

Question 36.
Explain a buffer action of an acidic buffer.
Answer:
Mechanism of action of an acidic buffer :
(1) An acidic buffer is a mixture of a weak acid and its salt with a strong base. The weak acid dissociates feebly, but the salt dissociates almost completely. Moreover, due to the common ions, largely supplied by the salt, dissociation of the weak acid is further suppressed.
(CH₃COOH + CH₃COONa) :
CH₃COONa_(aq) → CH₃COO_(aq) + \(\mathrm{Na}_{(\mathrm{aq})}^{+}\) (Complete)

(2) When a small quantity of strong acid (H⁺) is added to this mixture, hydrogen ions combine with acetate ions to form undissociated acetic acid. Thus, addition of an acid does not change the pH of the buffer.
\(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COOH}_{(\mathrm{aq})}\)
This removal of added H⁺ is called reserved basicity.

(3) When a small quantity of a strong base (OH^–) is added, the hydroxide ions react with the acid producing the corresponding anions and water. Thus, the concentrations of H⁺ and OH^– in the solution do not change and the pH remains constant.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{OH}_{(\mathrm{aq})}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
This removal of added OH^– is called reserved acidity.

Question 37.
Explain a buffer action of a basic buffer.
Answer:
Mechanism of action of a basic buffer :
(1) A basic buffer solution is a solution containing a weak base and its salt with a strong acid. The weak base dissociates feebly, but the salt dissociates completely. Moreover, due to the presence of the common ion, largely supplied by the salt, the dissociation of the base is further suppressed. (NH₄OH + NH₄Cl) :
\(\mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})} \rightarrow \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\) (Complete)

(2) When a small quantity of a strong acid is added to the solution, the hydrogen ions combine with the base producing corresponding cations and water. Thus, the addition of an acid does not change the pH of the buffer.
\(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
This removal of added H⁺ is called reserved basicity.

(3) When a small quantity of a strong base is added, the hydroxide ions combine with \(\mathrm{NH}_{4}^{+}\) ions to form undissociated NH₄OH. As a result, the hydrogen or hydroxyl ion concentration does not change. Thus, the pH of the solution does not change.
\(\mathrm{OH}_{(\mathrm{aq})}^{-}+\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
This removal of added OH^– is called reserved acidity.

Question 38.
What are properties of a buffer solution ?
OR
What are the advantages of a buffer solution ?
Answer:
Properties (or advantages) of a buffer solution :

1. The pH of a buffer solution is maintained appreciably constant.
2. By addition of a small amount of an acid or a base pH does not change.
3. On dilution with water, pH of the solution doesn’t change.

Question 39.
What are the applications of a buffer solution ?
Answer:
Buffer solutions have many applications as follows :
(1) In a biochemical system : Blood in our body has pH 7.36 – 7.42 due to (\(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2} \mathrm{CO}_{3}\)) and little change of 0.2 pH unit may be fatal. For example, saline solution used in intravenous injection contains a buffer solution maintaining pH of the blood in the required range.

(2) Agriculture : The properties of soil depend upon its pH. The salts present in soil such as phosphates, carbonates, bicarbonates and organic acids impart definite pH to the soil. Depending on pH the fertilizers are selected.

(3) Industry : In many industries, buffer solutions are used to carry out chemical processes very effectively, such as the industries of paper, dye, paints, drugs, ink, etc.

(4) Medicines : Many medicines particularly in the liquid state have a good stability and optimum activity at a definite pH, for which buffer solutions are used. For example penciline preparations are carried out in the presence of a buffer of sodium citrate. A buffer solution of magnesium citrate is prepared by adding citric acid to Mg(OH)₂.

(5) Analytical chemistry : In a qualitative analysis, the precipitation of groups, the chemical tests for detection of ions, etc. are carried out at a definite pH. For example, precipitation of cations of IIIA are carried in the presence of a basic buffer of pH 8 – 10 obtained by using NH₄OH and NH₄Cl.

Solved Examples 3.8

Question 40.
Solve the following :

(1) Calculate the pH of a buffer solution containing 0.1 M CH₃COOH and 0.05 M CH₃COONa. Dissociation constant of CH₃COOH is 1.8 × 10^-5 at 25 °C.
Solution :
Given : [CH₃COOH] = 0.1 M,
[CH₃COONa] = 0.05 M; K_a = 1.8 × 10^-5; pH = ?
pK_a = -log₁₀ K_a
= -log₁₀ 1.8 × 10^-5
= -(\(\overline{5} \cdot 2553\))
= 5 – 0.2553
= 4.7447
pH = \(\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\)
= \(4.7447+\log _{10} \frac{0.05}{0.1}\)
= 4.7447 + \((\overline{1} .6990)\)
= 4.7447 + (-1 + 0.6990)
= 4.7447 – 0.3010
= 4.4437
Ans. pH = 4.4437

(2) A buffer solution contains 0.3 M NH₄OH and 0.4 M NH₄Cl. If K_b for NH₄OH is 1.8 × 10^-5, calculate pH of the solution.
Solution :
Given : [NH₄OH] = 0.3 M; [NH₄Cl] = 0.4 M
K_b =1.8 × 10^-5; pH = ?
pK_b = -log₁₀ K_b
= -log₁₀ 1.8 × 10^-5
= \(-(\overline{5} .2553)\)
= 4.7447
pOH = \(\mathrm{p} K_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\mathrm{Base}]}\)
= \(4.7447+\log _{10} \frac{0.4}{0.3}\)
= 4.7447 + log₁₀ 1.333
= 4.7447 + 0.1248
= 4.8695
∵ pH + pOH = 14
∴ pH = 14 – pOH
= 14 – 4.8695
= 9.1305
Ans. pH = 9.1305.

(3) 0.2 dm³ acidic buffer solution contains 1.18 g acetic acid and 2.46 g sodium acetate. If K_a for acetic acid is 1.8 × 10^-5 at 25 °C, find pH of the solution.
Solution :

(4) A basic buffer solution contains 0.3 M NH₄OH and 0.2 M (NH₄)₂SO₄. If K_b for NH₄OH at a certain temperature is 2 × 10^-5, what is the pH of the solution ?
Solution :
Given : [NH₄OH] = 0.3 M
[NH₄⁺ ] = 2 × 0.2 = 0.4 M

pH + pOH = 14
∴ pH = 14 – pOH = 14 – 4.8239 = 9.1761
Ans. pH = 9.1761

Question 41.
Define solubility. How is it expressed ?
Answer:
Solubility : It is defined as the maximum amount of a substance in moles, that can be dissolved at constant temperature to give one litre of its saturated solution.
It is expressed in moles per litre or moles per decimeter cube of a saturated solution at given temperature.

Question 42.
Derive a relationship between solubility and solubility product.
Answer:
Consider a saturated solution of a spraingly soluble electrolyte (or salt) AxBy at a given constant temperature. Let S mol dm^-3 be the solubility of AxBy.
A following heterogeneous ionic equilibrium exists.

By a law of mass action, the equilibrium constant K will be represented as,

Since the active mass (concentration) of pure solid, A_xB_y(s) is treated as constant, [A_xB_y(s)] = K’
K × [A_xB_y(s)] = K × K’ = K_(sp)
Therefore, K_sp = [A^y+]^x × [B^x-]^y
where K_sp is called solubility product of A_xB_y.
At equilibrium the concentrations are,
[A^y+] = xS mol dm^-3
[B^x-]^y = yS mol dm^-3
∴ K_sp = [A^y+]^x × [B^x-]^y
= (xS)^x × (yS)^y
∴ K_sp = x^x.y^y.(S)^x+y ……….(1)
Hence solubility S is given by,
S = \(\left(\frac{K_{(\mathrm{sp})}}{x^{x} \cdot y^{y}}\right)^{\frac{1}{x+y}} \mathrm{~mol} \mathrm{dm}^{-3}\) …………(2)
The above equations, (1) and (2) give the relationship between solubility and solubility product.
Here x and y represent number of cations and anions respectively from the electrolyte.

Question 43.
Write expression for solubility and solubility product of following sparingly soluble salts : (1) AgBr (2) PbI₂ (3) Al(OH)₃
Answer:
In general, for a sparingly soluble salt A_xB_y,
K_sp = x^x.y^y.(S)^x+y

Question 44.
What is ionic product?
Answer:
Ionic product (IP) : It is defined as the product of concentrations in mol dm^-3 of ions of an electrolyte in the solution and denoted by IP.
In a saturated solution,
IP = K_sp where K_sp is the solubility product of the electrolyte.

Solved Examples 3.9

Question 45.
Solve the following :

(1) The solubility of AgBr in water is 1.28 × 10^-5 mol/dm³ at 298 K. Calculate the solubility product of AgBr at the same temperature.
Solution :
Given : S = 1.28 × 10^-5 mol dm^-3; K_sp = ?
AgBr dissociates as,
\(\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Br}_{(\mathrm{aq})}^{-}\)
K_sp = [Ag⁺] [Br^–]
As the solubility of AgBr in water is 1.28 × 10^-5 moles/dm³,
[Ag⁺] = [Br^–] = 1.28 × 10^-5 mol dm³
∴ K_sp = [1.28 × 10^-5] [1.28 × 10^-5]
= 1.638 × 10^-10
Ans. Solubility product of AgBr = 1.638 × 10^-10.

(2) The solubility of lead sulphate is 3.03 × 10^-5 kg/dm³. Calculate its solubility product. [Molecular mass of PbSO₄ = 303]
Solution :
Given : S = 3.03 × 10^-5 kg dm^-3, K_sp = ?
Lead sulphate dissociates as
\(\begin{aligned}
&\mathrm{PbSO}_{4} \rightleftharpoons \mathrm{Pb}^{2+}+\mathrm{SO}_{4}^{2-} \\
&\text { (solid) } \quad \text { (aq) } \quad \text { (aq) }
\end{aligned}\)
Molecular weight of PbSO₄ = 303
= 303 × 10^-3 kg
The solubility of PbSO₄ is 3.03 × 10^-5 kg/dm³.
Solubility in mol dm^-3
Weight of PbSO₄ per dm³ Molecular weight 3.03 × 10^-5

(3) The solubility product of AgBr is 3.3 × 10^-12 at 298 K. What concentration of Br^– ion is needed to precipitate AgBr from solution of 0.01 M Ag⁺?
Solution :
Given : K_sp = 3.3 × 10; [Ag⁺] = 1 × 10^-2 M;
[Br^–] = ?
AgBr dissociates as
\(\begin{aligned}
&\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-} \\
&\text {(solid) } \quad \text { (aq) } \quad \text { (aq) }
\end{aligned}\)
K_sp = [Ag⁺] [Br^–]
K_sp = Solubility product = 3.3 × 10^-12
[Ag⁺] = Concentration of Ag⁺
= 0.01 = 1.0 × 10^-2 M
[Br^–] = Concentration of Br^– = ?
∴ 3.3 × 10^-12 = 1.0 × 10^-2 × [Br^–]
∴ [Br^–] = 3.3 × 10^-10 mol/dm³
Ans. The concentration of Br^– required for precipitation of AgBr should be greater than 3.3 × 10^-10 mol/dm³.

(4) The solubility product of magnesium hydroxide is 1.4 × 10^-11. Calculate the solubility of magnesium hydroxide.
Solution :
Given : K_sp =1.4 × 10^-11; S = ?
Magnesium hydroxide dissociates as shown below :
Mg(OH)₂ ⇌ Mg²⁺ + 2(OH)^–
K_sp = [Mg²⁺] [OH^–]²
Let the solubility of Mg(OH)₂ be S mol dm^-3.
∴ [Mg²⁺] = Concentration of Mg²⁺ ions
= S mol dm^-3
∴ [OH^–] = Concentration of OH^– ions
= 2S mol dm^-3
∴ K_sp = S × (25)² = 4S³
K_sp = 1.4 × 10^-11
∴ 1.4 × 10^-11 =4S³
1.4 × 10^-11 = 4S³.

S = 1.518 × 10^-4 mol dm^-3
Ans. Solubility of Mg(OH)₂
= 1.518 × 10^-4 mol dm^-3

(5) The solubility of silver chloride is 1.562 × 10^-10 mol dm^-3 at 298 K. Find its solubility in g dm^-3 at the same temperature.
Solution :
Given :
Solubility of AgCl = S = 1.562 × 10^-10 mol dm^-3
Solubility of AgCl in g dm^-3 = ?
Molar mass of AgCl = M = 143.5 g mol^-1
Solubility in gram per dm³
= solubility in mol dm^-3 × molar mass
= 1.562 × 10^-10 × 143.5
= 2.241 × 10^-5 g dm-3
Ans. Solubility of AgCl = 2.241 × 10^-8 g dm^-3

(6) The solubility of PbSO₄ in water is 0.038 g dm^-3 at room temperature. Calculate its solubility and solubility product at the same temperature. (Atomic weights : Pb = 207.3, S = 32, O = 16)
Solution :
Given : Solubility of PbSO₄ = 0.038 g/dm^-3
Molar mass of PbSO₄ = 303.3 g mol^-1
Solubility in mol dm^-3 = ?
K_sp =?
Solubility in mol dm^-3 = \(\frac{0.038}{303.3}\)
= 1.253 × 10^-4 mol dm^-3

Ans. S = 1.253 × 10^-4 mol dm^-3;
K_sp = 1.57 × 10^-8

(7) The solubility product of PbS at 298 K is 4.2 × 10^-28, The concentration of Pb⁺⁺ ion is 0.001 M. Calculate S^2- ion concentration at which PbS just gets precipitated.
Solution :
Given :
Solubility product of PbS = K_sp = 4.2 × 10^-28
Concentration of Pb⁺⁺ = [Pb⁺⁺] = 0.001 M
Concentration of S^{– –} = [S^{– –}] = ?
For PbS,
\(\mathrm{PbS}_{(\mathrm{s})} \rightleftharpoons \mathrm{Pb}^{++}+\mathrm{S}^{–}\)
∴ K_sp = [Pb⁺⁺] × [S^{– –}]
∴ [S–] = \(\frac{K_{\mathrm{sp}}}{\left[\mathrm{Pb}^{++}\right]}\)
= \(\frac{4.2 \times 10^{-28}}{0.001}\)
= 4.2 × 10^-25 M
To precipitate Pb⁺⁺ as PbS, ionic product must be greater than 4.2 × 10^-28.
Hence, [S^{– –}] > 4.2 × 10^-25 M.
Ans. Concentration of S^{– –} required > 4.2 × 10^-25 M

(8) At 298 K, the solubility of silver sulphate is 1.85 × 10^-2 mol dm^-3. Calculate the solubility product of silver sulphate.
Solution :
Given : Silver sulphate dissociates as follows :

Solubility of Ag₂SO₄ = 1.85 × 10^-2 mol dm^-3
K_sp = Solubility product of Ag₂SO₄ = ?
[Ag⁺] = Concentration of Ag⁺ ion
= 2 × 1.85 × 10^-2 = 3.70 × 10^-2 mol dm^-3
[latex]\mathrm{SO}_{4}^{2-}[/latex] = Concentration of \(\mathrm{SO}_{4}^{2-}\)
= 1.85 × 10^-2 mol dm^-3
∴ K_sp = (3.70 × 10^-2)² × (1.85 × 10^-2)
= 13.69 × 10^-4 × 1.85 × 10^-2
= 25.33 × 10^-6
= 2.533 × 10^-5
Ans. Solubility product of Ag₂SO₄
= 2.533 × 10^-5

Question 46.
What is common ion?
Answer:
Common ion : An ion common to two electrolytes is called common ion. This is generally applicable to a mixture of a strong and a weak electrolyte. For example, a solution containing weak electrolyte CH₃COOH and strong electrolyte salt CH₃COONa.
CH₃COONa → CH₃COO^– + Na⁺;
CH₃COOH ⇌ CH₃COO^– + H⁺
Hence CH₃COOH and CH₃COONa have a common ion CH₃COO^–.

Question 47.
Define the term common ion effect.
Answer:
Common ion effect : The suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte having an ion in common with the weak electrolyte is called common ion effect. For example, CH₃COOH and CH₃COONa have common ion CH₃COO^–.

Question 48.
Explain common ion effect with suitable example.
Answer:
A weak electrolyte dissociates partially in aqueous solution to produce cations and anions. Equilibrium exists between ions thus formed and the undissociated molecules.
BA ⇌ B⁺ +A^–
For such an equilibrium, the dissociation constant K is defined as
K = \(\frac{\left[\mathrm{B}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{BA}]}\)
K is constant for the weak electrolyte at a given temperature.
Now, if another electrolyte BC or DA is added to the solution BA, having a common ion either B⁺ or A^–, then the concentration of either B⁺ or A^– is increased. However, as K is always constant, the increase in the concentration of any one of the ions shifts the equilibrium to left. In other words, the dissociation of BA is suppressed. This is called common ion effect. For example, the dissociation of a weak acid CH₃COOH is suppressed by adding CH₃COONa having common ion CH₃COO^–.
CH₃COOH ⇌ CH₃COO^– + H⁺
CH₃COONa → CH₃COO^– + Na⁺

Question 49.
Explain the common ion effect on dissociation of a weak acid.
Answer:
(1) Consider the dissociation or ionisation of a weak acid, CH₃COOH in its solution.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+}\)
The dissociation constant Ka for CH₃COOH will be,
K_a = \(\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \times\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\)
K_a is constant for CH₃COOH at constant temperature.

(2) If a strong electrolyte like salt CH₃COONa is added to the solution of CH₃COOH, then on dissociation it gives a common ion CH₃COO^–.
CH₃COONa → CH₃COO^– + Na⁺

(3) Due to common ion CH₃COO^–, overall concentration of CH₃COO^– in the solution is increased, which increases the ratio,
\(\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \times\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\). In order to keep this ratio constant, the concentration of H⁺ is decreased, by shifting the equilibrium to the left hand side according to Le Chatelier’s principle.

(4) Thus the ionisation of a weak acid is suppressed by a common ion.

Question 50.
Explain the effect of common ion on the dissociation of weak base.
Answer:
(1) Consider the dissociation or ionisation of a weak base, NH₄OH in its dilute solution. \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The dissociation constant K_b for NH₄OH will be,
\(K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}\)

(2) If a strong electrolyte like salt NH₄Cl is added to the solution of NH₄OH, then it gives common ion \(\mathrm{NH}_{4}^{+}\).
NH₄Cl → \(\mathrm{NH}_{4}^{+}\) + Cl^–

(3) Due to common ion \(\mathrm{NH}_{4}^{+}\), overall concentration of \(\mathrm{NH}_{4}^{+}\) is increased, which increases the ratio \(\left[\mathrm{NH}_{4}^{+}\right]\) × [OH^–]/[NH₄OH],
In order to keep this ratio constant, the equilibrium is shifted to the left hand side which satisfies Le Chatelier’s principle.

(4) Thus the ionisation of a weak base is suppressed by a common ion.

Multiple Choice Questions

Question 51.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. According to Lowry-Bronsted concept, base is a substance which acts as –
(a) a proton donor
(b) an electron donor
(c) a proton acceptor
(d) an electron acceptor
Answer:
(c) a proton acceptor

2. BF₃ is a
(a) Lewis acid
(b) Lewis base
(c) amphoteric compound
(d) Electrolyte only
Answer:
(a) Lewis acid

3. Which of the following is a conjugate acid-base pair ?
(a) HCl, NaOH
(b) KCN, HCN
(c) NH₄Cl, NH₄OH
(d) H₂SO₄, \(\mathrm{HSO}_{4}^{-}\)
Answer:
(d) H₂SO₄, \(\mathrm{HSO}_{4}^{-}\)

4. The conjugate acid of \(\mathrm{NH}_{2}^{-}\) is
(a) NH₃
(b) NH₂OH
(c) \(\mathrm{NH}_{4}^{+}\)
(d) N₂H₄
Answer:
(a) NH₃

5. Which of the following molecules is not a Lewis base?
(a) H₂O
(b) BF₃
(c) NH₃
(d) CO
Answer:
(b) BF₃

6. In the following reaction
\(\mathrm{HC}_{2} \mathrm{O}_{4(\mathrm{aq})}^{-}+\mathrm{PO}_{4}^{3-} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{C}_{2} \mathrm{CO}_{4}^{2-}\) Which of two are Lowry-Bronsted bases ?
(a) \(\mathrm{HC}_{2} \mathrm{C}_{4}^{-} \text {and } \mathrm{PO}_{4}^{3-}\)
(b) \(\mathrm{HPO}_{4}^{2-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)
(c) \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \text {and } \mathrm{HPO}_{4}^{2-}\)
(d) \(\mathrm{PO}_{4}^{3-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)
Answer:
(d) \(\mathrm{PO}_{4}^{3-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)

7. According to the Arrhenius theory,
(a) an acid is a proton donor
(b) an acid is an electron pair acceptor
(c) a hydrogen ion exists freely in an aqueous solution
(d) a hydrogen ion is always hydrated to form a hydrogen ion
Answer:
(c) a hydrogen ion exists freely in an aqueous solution

8. The species which will behave both as a conjugate acid and base is
(a) NH₄OH
(b) \(\mathrm{CO}_{3}^{–}\)
(c) \(\mathrm{HSO}_{4}^{-}\)
(d) H₂SO₄
Answer:
(c) \(\mathrm{HSO}_{4}^{-}\)

9. According to the Lewis theory, an acid is
(a) nucleophile
(b) an electrophile
(c) a proton acceptor
(d) an electron donor
Answer:
(b) an electrophile

10. If a 0.1 M solution of HCN is 0.01% dissociated, the dissociation constant for HCN is,
(a) 10^-3
(b) 10⁺³
(c) 10^-7
(d) 10^-9
Answer:
(d) 10^-9

11. The pH of decimolar solution KOH is
(a) 1
(b) 4
(c) 10
(d) 13
Answer:
(d) 13

12. Ostwald’s dilution law is applicable in case of dilute solution of
(a) HCl
(b) H₂SO₄
(c) NaOH
(d) CH₃COOH
Answer:
(d) CH₃COOH

13. The degree of dissociation of a 0.1 M monobasic acid is 0.4%. Its dissociation constant is
(a) 0.4 × 10^-4
(b) 4.0 × 10^-4
(c) 1.6 × 10^-6
(d) 0.8 × 10^-5
Answer:
(c) 1.6 × 10^-6

14. The ionic product of water will increase, if
(a) Pressure is decreased
(b) H⁺ ions are added
(c) OH^– ions are added
(d) Temperature is increased
Answer:
(d) Temperature is increased

15. The [OH^–] for a weak base of dissociation constant K_b and concentration C is nearly equal to
(a) \(\sqrt{\frac{K_{\mathrm{b}}}{C}}\)
(b) KbC
(c) \(\sqrt{K_{\mathrm{b}} C}\)
(d) \(\frac{C}{K_{\mathrm{b}}}\)
Answer:
(c) \(\sqrt{K_{\mathrm{b}} C}\)

16. The [H⁺] for a weak acid of dissociation constant K_a and concentration C is nearly equal to
(a) \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
(b) \(\sqrt{K_{\mathrm{a}} C}\)
(c) \(\frac{K_{\mathrm{a}}}{\sqrt{c}}\)
(d) \(\frac{c}{K_{\mathrm{a}}}\)
Answer:
(b) \(\sqrt{K_{\mathrm{a}} C}\)

17. Which of the following solution with same concentration will have highest pH
(a) Al(OH)₃
(b) K₂CO₃
(c) NH₄OH
(d) NaOH
Answer:
(d) NaOH

18. 10 ml of 0.1 M H₂SO₄ is mixed with 20 ml of 0.1 M KOH, the pH of resulting solution will be
(a) 0
(b) 7
(c) 2
(d) 9
Answer:
(b) 7

19. The gastric juice in our stomach contains enough hydrochloric acid to make the hydrogen ion concentration 0.01 mol/dm³. The pH of gastric juice is-
(a) 0.01
(b) 1
(c) 2
(d) 14
Answer:
(c) 2

20. If the hydrogen ion concentration of an acid is decreased ten times, its pH will be
(a) increased by one
(b) decreased by one
(c) remains unchanged
(d) increase by 10
Answer:
(a) increased by one

21. Which of the following metal sulphide is precipitated in an acidic medium ?
(a) NiS
(b) CoS
(c) CuS
(d) MnS
Answer:
(c) CuS

22. The relationship between the solubility and solubility product for silver carbonate is
(a) K_sp = s²
(b) \(\sqrt{K_{\mathrm{sp}}}\) = 4s²
(c) K_sp = 27S⁴
(d) K_sp = 4s³
Answer:
(d) K_sp = 4s³

23. If ‘S’ is solubility in mol dm^-3 and K_sp is solubility product of BA₂ type of salt, then relation between them is
(a) S = \(\sqrt{K_{\mathrm{sp}}}\)
(b) K_sp = 4S³
(c) K_sp = S³
(d) S = K_sp
Answer:
(b) K_sp = 4S³

24. The addition of solid sodium carbonate to pure water results in
(a) an increase in H⁺ ion concentration
(b) an increase in pH
(c) no change in pH
(d) a decrease in OH^– concentration
Answer:
(b) an increase in pH

25. Which of the following salt, when dissolved in water will hydrolyse ?
(a) NaCl
(b) NH₄Cl
(c) KCl
(d) Na₂SO₄
Answer:
(b) NH₄Cl

26. A solution of blue vitriol is acidic in nature because
(a) CuSO₄ reacts with water
(b) Cu²⁺ ions reacts with water
(c) SO₄^2- ions reacts with water
(d) CuSO₄ removes OH^– ions from water
Answer:
(a) CuSO₄ reacts with water

27. What is the nature of the solution of salt FeCl₃ ?
(a) Acidic
(b) Basic
(c) Neutral
(d) Amphoteric
Answer:
(a) Acidic

28. Which of the following salts does not hydrolyse in water ?
(a) Sodium acetate
(b) Sodium carbonate
(c) Sodium nitrate
(d) Sodium cyanide
Answer:
(c) Sodium nitrate

29. An aqueous solution of magnesium chloride changes blue litmus red due to
(a) the formation of Cl^– ions
(b) the formation Mg²⁺ ions
(c) reaction of Cl^– ions with water
(d) hydrolysis of the salt
Answer:
(d) hydrolysis of the salt

30. An aqueous solution of which of the following salts is basic ?
(a) CH₃COONa
(b) NH₄Cl
(c) KNO₃
(d) CuSO₄
Answer:
(a) CH₃COONa

31. The number of moles of hydroxide ions (OH^–) produced from 2 moles of Na₂CO₃ is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

32. The POH value for solution is 4, its hydrogen ion concentration will be
(a) 10^-4
(b) 10^-10
(c) 10¹⁰
(d) 10⁴
Answer:
(b) 10^-10

33. If an acid is diluted
(a) pH increases
(b) pH decreases
(c) no change occurs
(d) can vary depending on an acid
Answer:
(a) pH increases

34. pH of a solution is 13. H⁺ ions present in 1 cm³ of the solution is
(a) 6.023 × 10¹⁰
(b) 6.023 × 10⁷
(c) 6.023 × 10^-10
(d) 6.023 × 10^-7
Answer:
(b) 6.023 × 10⁷

35. pH of blood is maintained constant by mechanism of
(a) common ion effect
(b) buffer
(c) solubility
(d) all of these
Answer:
(b) buffer

36. The pH of 0.05 M solution of dibasic acid is
(a) +1
(b) -1
(c) +2
(d) -2
Answer:
(a) +1

37. The pH of a 0.63% nitric acid solution is (Equivalent weight of nitric acid is 63)
(a) 6
(b) 7
(c) 1
(d) 9
Answer:
(c) 1

38. 100 ml of 0.01 M solution of NaOH is diluted to 1 dm³. What is the pH of the dilute solution?
(a) 12
(b) 11
(c) 2
(d) 3
Answer:
(b) 11

39. If the H⁺ ion concentration in a solution is 0.01 M, the pOH of the solution is
(a) 12
(b) 10^-10
(c) 2
(d) 14
Answer:
(a) 12

40. If the pH value of a solution is zero, the solution is
(a) a strong acid
(b) a very weak acid
(c) neutral
(d) a base
Answer:
(a) a strong acid

41. The pH of a solution is 5, when the hydroxyl ion concentration is
(a) 10^-5 mol/dm³
(b) 10^-7 mol/dm³
(c) 10^-9 mol/dm³
(d) 10^-14 mol/dm³
Answer:
(c) 10^-9 mol/dm³

42. The pH of human blood in a normal person is approximately
(a) 4.7
(b) 6.04
(c) 7.40
(d) 8.74
Answer:
(c) 7.40

43. If molarity of NaOH is 3.162 × 10^-3 M, its pH is
(a) 8.5
(b) 9.5
(c) 10.5
(d) 11.5
Answer:
(d) 11.5

44. The common ion effect is based on
(a) Sorensen’s principle
(b) Le Chatelier’s principle
(c) Heisenberg’s principle
(d) Freundlich’s principle
Answer:
(b) Le Chatelier’s principle

45. The ion that cannot be precipitated by both HCl and H2S is
(a) Pb²⁺
(b) Cu²⁺
(c) Ag⁺
(d) Ca²⁺
Answer:
(d) Ca²⁺

46. The correct representation for solubility product of SnS2 is
(a) [Sn⁴⁺] [S^2-]²
(b) [Sn⁴⁺] [S^2-]
(c) [Sn⁴⁺] [2S^2-]
(d) [Sn⁴⁺] [2S^2-]
Answer:
(a) [Sn⁴⁺] [S^2-]²

47. The solubility product of a salt BA at room temperature is 1.21 × 10^-6. Its molar solubility is
(a) 1.21 × 10^-3 M
(b) 1.1 × 10^-4 M
(c) 1.1 × 10^-3 M
(d) 1.21 × 10^-2 M
Answer:
(c) 1.1 × 10^-3 M

48. Among the following hydroxides, the one which has the lowest value of solubility product at temperature 298 K is,
(a) Mg(OH)₂
(b) Ca(OH)₂
(c) Ba(OH)₂
(d) Be(OH)₂
Answer:
(d) Be(OH)₂

49. A solution becomes unsaturated when
(a) ionic product = solubility product
(b) ionic product < solubility product
(c) ionic product > solubility product
(d) ionic product ≥ solubility product
Answer:
(b) ionic product < solubility product

50. The solubility product of Fe(OH)₃ is
(a) [latex]\mathrm{F}_{\mathrm{e}}^{2+}[/latex] [OH^–]³
(b) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]²
(c) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]³
(d) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex]³ [OH^–]³
Answer:
(c) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]³

51. The solubility product of PbS in 4.2 × 10^-28 at 300 K. The sulphide ions concentration required to precipitate PbS from a solution containing 0.001 M of lead ion is
(a) ≥ 2.1 × 10^-14 mol/dm³
(b) ≥ 4.2 × 10^-14 mol/dm³
(c) ≥ 4.2 × 10^-25 mol/dm³
(d) ≤ 4.2 × 10^-28 mol/dm³
Answer:
(c) ≥ 4.2 × 10^-25 mol/dm³

52. 0.025 M CH₃COOH is dissociated 9.5%. Hence the pH of the solution is
(a) 2.6244
(b) 3.128
(c) 2.988
(d) 2.267
Answer:
(a) 2.6244

53. 0.1 M HCN is dissociated 0.01%. The dissociation constant of HCN is
(a) 1.1 × 10^-6
(b) 1 × 10^-8
(c) 1 × 10^-9
(d) 1 × 10^-7
Answer:
(c) 1 × 10^-9

54. The solubility of product of a sparingly soluble salt AB₂ is 3.2 × 10^-11. If solubility in mol dm^-3 is
(a) 4 × 10^-4
(b) 3.2 × 10^-4
(c) 1 × 10^-5
(d) 2 × 10^-4
Answer:
(d) 2 × 10^-4

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions

Question 1.
What is solution?
Answer:
The solution is a homogeneous mixture of two or more components or pure substances. When the size of particles of the components is of the order of 10^-10 m, then the solution is called a true solution. E.g. An aqueous solution of sugar.

Question 2.
Explain : (1) Homogeneous solution
(2) Heterogeneous solution.
Answer:
(1) Homogeneous solution : A solution in which solute and solvent form uniform homogeneous one phase due to attraction between their molecules/particles is called homogeneous solution.
E.g. A solution of NaCl or sugar.

(2) Heterogeneous solution : A solution consisting of two or more phases is called a heterogeneous solution. E.g. A colloidal solution of starch.

Question 3.
Define : (1) Solvent (2) Solute.
Answer:
(1) Solvent : The component in which solution formation takes place and which constitutes larger proportion of a solution is called solvent. For example, in an aqueous solution of sugar, water is the solvent.

(2) Solute : In a solution the component which dissolves and constitutes smaller proportion of a solution is called a solute. For example, in a sugar solution, sugar is the solute.

Question 4.
What are the different types of solutions?
Answer:
A solution consists of a solvent and a solute. Since the physical states of a solvent and a solute may be gaseous, liquid or a solid, there are nine types of solutions.

Question 5.
Mention the solvent and solute in the following :
(1) Smoke
(2) Moisture
(3) Alloy
(4) Soda water.
Answer:

Question 6.
What is a saturated solution?
Answer:
A solution which contains the maximum amount of dissolved solute and further the solute can’t be dissolved is called a saturated solution.

There exists a dynamic equilibrium which is represented as,

Question 7.
What is a supersaturated solution?
Answer:
A solution containing a solute more than that required to form a saturated solution at equilibrium is called a supersaturated solution.

When a tiny crystal of a solute is added to supersaturated solution, the excess solute separates out and forms saturated solution.

Question 8.
What is solubility of a solute ?
Answer:
Solubility : It is defined as amount of a solute present per unit volume in its saturated solution at a specific temperature.
It is expressed in mol L^-1 or mol dm^-3.

Question 9.
Explain the factors on which the solubility of a substance (solute) depends.
Answer:
The extent of dissolution of a substance (solute) depends upon the following factors :
(1) Nature of a solute : A solute may be crystalline, amorphous, ionic or covalent. Hence accordingly its tendency to dissolve changes. The substances having similar intermolecular forces tend to dissolve in each other.

(2) Nature of solvents : Solvents are classified as polar and nonpolar. Polar solutes dissolve in polar solvents. For example, ionic compounds dissolve in polar solvent like water. A solvent other than water is called a nonaqueous solvent. For example, C₆H₆, CCl₄, etc. Solutions in these solvents are called nonaqueous solutions.
The solvent may be a gas, a liquid or a solid.

(3) Amount of a solvent : More amount of a solvent, will dissolve more quantity of the solute.

(4) Temperature : Depending on the nature of a solvent and a solute the solubility changes with termperature. The effect depends on the heat of solution, hydration energy, etc. Generally as the temperature increases, solubility of solid increases and that of gases decreases.

(5) Pressure :

• Pressure has no effect on the solubilities of solids and liquids since they are incompressible.
• The effect of pressure is important only for solutions which involve gases as solutes. With the increase in pressure and decrease in temperature, the solubility of gases increases.

Question 10.
Explain with the help of Le Chatelier’s principle the effect of temperature on solubility.
Answer:

1. The effect of temperature on solubility depends on enthalpy of solution.
2. For example, dissolution of KCl in water is an endothermic process since heat is absorbed during dissolution. In according to Le Chatelier’s principle by increasing temperature the solubility of KCl increases.
3. Dissolution of CaCl₂, Li₂SO₄, H₂O in water is an exothermic process since heat is evolved during dissolution. In this, according to Le Chatelier’s principle by increasing the temperature the solubility decreases.

Question 11.
Explain the solubility of gases in liquids.
Answer:

• Gases are soluble in water and other liquids and their solubility depends upon the nature of the gas.
• Non-polar gases like O₂, have less solubility in polar solvents.
• Polar gases like CO₂, NH₃, HCl, etc. are more soluble in polar solvent like water. CO₂ forms H₂CO₃, while NH₃ forms NH₄OH in aqueous solutions.
• The solubility of gases in liquids increases with the increase in pressure and the decrease in temperature.

Question 12.
Why does the solubility of a gas decrease with increase in temperature ?
OR
How does solubility of a gas in water varies with temperature ?
Answer:

• The gases are soluble in water and other liquids.
• According to Charles’ law, the volume of a given mass of a gas increases with the increase in temperature at constant pressure.
• Hence, the volume of the dissolved gas increases with the increase in temperature.
• This enormous increase in volume of the gases cannot be accommodated by the solvent molecules, hence excess of the gases escape out in the form of bubbles.

Therefore, the solubility of gases in liquids decreases with temperature.

Question 13.
Explain the effect of pressure on the solubility of the gases.
OR
State and explain Henry’s law.
Answer:
(1) Since the gases are compressible, their solubility in the liquids is influenced by external pressure of the gas. The solubility of gases increases with the increase in pressure.

(2) Henry’s law : It states that the solubility of a gas. in a liquid at constant temperature is proportional to the pressure of the gas above the solution.
(i) If S is the solubility of a gas in mol dm^-3 at a pressure P and constant temperature then by Henry’s law,
S ∝ P or S = K_H × P
where K_H is called Henry’s law constant.
(ii) If P = 1 atm, then S = K_H.
(iii) If several gases are present, then the solubility of any gas in the mixture is proportional to its partial pressure at given temperature.

(3) Illustration of Henry’s law : In case of aerated or carbonated drink beverage, the bottle is filled by dissolving CO₂ gas at high pressure and then sealed.
Above the liquid surface there is air and undissolved CO₂. Due to high pressure, the amount of dissolved CO₂ is large.

When the cap of the aerated bottle is removed, the pressure on the solution is lowered, hence excess of CO₂ and air escape out in the form of effervescence. Thus by decreasing the pressure, solubility of CO₂ is decreased.

Question 14.
What are the exceptions to Henry’s law? Why ?
Answer:
(1) The gases like NH₃ and CO₂ do not obey Henry’s law.
(2) This is because, these gases react with water,
\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \longrightarrow \mathrm{NH}_{4 \text { (aq) }}^{+}+\mathrm{OH}_{(\text {aq })}^{-}\)
CO_2(g) + H₂O(l) → H₂CO_3(aq)
(3) Due to reactions of the gases like NH₃, CO_2(g), they have higher solubilities than expected by Henry’s law.

Question 15.
Oxygen gas is slightly soluble in water but it is highly soluble in blood. Explain.
Answer:
The vital constituent of blood, namely haemoglobin reacts with oxygen increasing the solubility of oxygen.
Haemoglobin + 4O_2(g) → Haemoglobin O₈
This oxygenated blood is circulated to the various parts of body, for the supply of oxygen.

Question 16.
Obtain the units of Henry’s law constant.
Answer:
By Henry’s law, S = K_H × P, where S is solubility of the gas in mol dm^-3, P is the pressure of the gas in atmosphere (or in bar) and K_H is Henry’s law constant.
∴ K_H = \(\frac{S^{\left(\mathrm{mol} \mathrm{dm}^{-3}\right)}}{P_{(\mathrm{atm})}}\) mol dm^-3 atm^-1 or mol dm^-3 bar^-1
Hence the units of Henry’s law constant K_H are mol dm^-3atm^-1.

Solved Examples 2.4

Question 17.
Solve the following :
(1) For a gas, the Henry’s law constant is 1.25 × 10^-3 mol dm^-3 atm^-1 at 25 °C. Calculate the solubility of the given gas at 2.5 atm and 25 °C.
Solution :
Given : Henry’s law constant = K_H
= 1.25 × 10^-3 mol dm^-3 atm^-1
Pressure of the gas = P = 2.5 atm
Solubility of the gas = S = ?
By Henry’s law
S = K_H × P
= 1.25 × 10^-3 mol dm^-3 atm^-1 × 2.5 atm
= 3.125 × 10^-3 mol dm^-3
Ans. Solubility of gas = 3.125 × 10^-3 mol dm^-3

(2) The solubility of dissolved oxygen to 27 °C is 2.6 × 10^-3 mol dm^-3 at 2 atm. Find its solubility at 8.4 atm and 27 °C.
Solution :
Given : Solubility of O₂
= S₁ = 2.6 × 10^-3 mol dm^-3
Initial pressure of O₂ = P₁ = 2 atm
Final pressure of O₂ = P₂ = 8.4 atm
Solubility of O₂ = S₂ = ?
(i) By Henry’s law,
S₁ = K_H × P₁
∴ Henry’s law constant K_H is,
K_H = \(\frac{S_{1}}{P_{1}}=\frac{2.6 \times 10^{-3}}{2}\)
= 1.3 × 10^-3 mol dm^-3 atm^-1
(ii) Now, S₂ = K_H × P₂ = 1.3 × 10^-3 × 8.4
= 10.92 × 10^-3
= 1.092 × 10^-2 mol dm^-3
Ans. Solubility of O₂ = 1.092 × 10^-2 mol dm^-3

(3) Henry’s law constant for the solubility of methane in benzene is 4.27 × 10^-5 mm^-1 Hg mold dm^-3 at constant temperature. Calculate the solubility of methane at 760 mm Hg pressure at same temperature.
Solution :
Given : Henry’s law constant = K_H
= 4.27 × 10^-5mm^-1 Hg mol dm^-3
Pressure of the gas = P = 760 mm Hg
K_H = 4.27 × 10^-5 mm^-1 mol dm^-3
= 4.27 × 10^-5 × 760 atm^-1 mol dm^-3
= 3245 × 10^-5 atm^-1 mol dm^-3
= 3.245 × 10^-2 atm^-1 mol dm^-3
P = 760 mm = \(\frac{760}{760}\) atm = 1 atm
By Henry’s law,
S = K_H × P = 3.245 × 10^-2 atm^-1 mol dm^-3 × 1 atm
= 3.245 × 10^-2 mol dm^-3
Ans. Solubility of methane = 3.245 × 10^-2 mol dm^-3

(4) The solubility of ethane at 25 °C is 0.92 × 10^-3 g dm^-3 at 1000 mm Hg pressure. Calculate Henry’s law constant.
Solution :
Given : Solubility of ethane
= S = 0.92 × 10^-5 g dm^-3
Pressure of ethane = P = 1000 mm Hg
Molar mass of ethane (C₂H₆) = 30 g mol^-1
Henry’s law constant = K_H = ?
S = 0.92 × 10^-3 g dm^-3
= \(\frac{0.92}{30}\) × 10^-3
= 3.067 × 10^-5 mol dm^-3
P = 1000 mm = \(\frac{1000}{760}\)atm = 1.316 atm
By Henry’s law,
S = K_H × P
∴ K_H = \(\frac{S}{P}=\frac{3.067 \times 10^{-5}}{1.316}\)
= 2.33 × 10^-5 mol dm^-3 atm^-1
Ans. Henry’s law constant = K_H
= 2.33 × 10^-5 mol dm^-3 atm^-1

(5) The solubility of nitrogen at 30 °C is 2.5 × 10^-3 g dm^-3 at 760 mm pressure. What will be its solubility in mol dm^-3 at 20,000 mm and same temperature?
Solution :
Given : Initial solubility of N₂ = S₁ =2.5 × 10^-3 g dm^-3
Initial pressure = P₁ = 760 mm
Final pressure = P₂ = 20,000 mm
Final solubility = S₂ in mol dm^-3 = ?
Molar mass of N₂ gas = 28 g mol^-1
S₁ = 2.33 × 10^-5 mol dm^-3
= \(\frac{2.5 \times 10^{-3}}{28}\) mol dm^-3
= 8.93 × 10^-5 mol dm^-3
P₁ = \(\frac{760}{760}\) = 1 atm
P₂ = \(\frac{20,000}{760}\) = 26.32 atm
By Henry’s law,
S₁ = K_H × P₁
∴ K_H = \(\frac{S_{1}}{P_{1}}=\frac{8.93 \times 10^{-5}}{1}\)
= 8.93 × 10^-5 mol dm^-3 atm^-1
S₂ = K_H × P₂ = 8.93 × 10^-5 × 26.32
= 2.35 × 10^-3 mol dm^-3
Ans. Solubility of N₂ gas
= 2.35 × 10^-3 mol dm^-3.

[Alternative method:
S₁ = K_HP₁ and S₂ = K_HP₂
∴ \(\frac{S_{2}}{S_{1}}=\frac{K_{\mathrm{H}} P_{2}}{K_{\mathrm{H}} P_{1}}=\frac{P_{2}}{P_{1}}\)
∴ S₂ = S₁ × \(\frac{P_{2}}{P_{1}}\) = 8.93 × 10^-5 × \(\frac{20,000}{760}\)
= 2.35 × 10^-3 mol dm^-3]

Question 18.
Marine life like fish prefers to stay at lower level in water. Explain.
OR
Explain, why do aquatic animals prefer to stay at lower level of water during summer?
Answer:

1. The solubility of oxygen gas decreases with the increase in temperature.
2. In sea or lake water, the temperature of upper level is higher than the lower level.
3. Therefore the dissolved oxygen content in water is more at lower level than at higher level required for the marine life.

Hence marine life like fish prefers to stay at lower level than upper level of water.

Question 19.
State and explain Raoult’s law.
Answer:
Statement of Raoult’s law : The law states that, at constant temperature, the partial vapour pressure of any volatile component of a solution is equal to the product of vapour pressure of the pure component and the mole fraction of that component in the solution.

Let P₀ and P be the respective vapour pressures of a pure volatile component and a solution. If x₁ is the mole fraction of a solvent then by Raoult’s law,
P = x₁ × P₀.

Explanation : Consider a solution containing two volatile components A and B having mole fractions x₁ and x₂ respectively.
Let \(P_{1}^{0}\) and \(P_{2}^{0}\) be the vapour pressures of pure components (or liquids) A and B respectively.
Then by Raoult’s law, vapour pressure of component A = P₁ = X₁ × \(P_{1}^{0}\), vapour pressure of component B = P₂ = x₂ × \(P_{2}^{0}\).
Here P₁ and P₂ represent partial vapour pressures of the two liquid components in the solution.
Hence the total vapour pressure, P_T of the solution will be,
P_T = P₁ + P₂
∴ P_T = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
∵ x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P_T = (1 – x₂)\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= \(P_{1}^{0}\) – x₂\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
With the help of above equation, the vapour pressures of solutions having different concentrations (or mole fractions) can be calculated.

Question 20.
Explain the variation of vapour pressure with mole fraction of a solute in a liquid mixture.
Answer:
Consider a liquid mixture of two liquid components A and B having vapour pressures \(P_{1}^{0}\) and \(P_{2}^{0}\) and mole fractions x₁ and x₂ respectively.
By Raoult’s law, the vapour pressures P₁ and P₂ are,
P₁ = x₁\(P_{1}^{0}\) and P₂ = x₂\(P_{2}^{0}\)
The vapour pressure of the solution is,
P_T = P₁ +P₂ = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)

∴ P_T = (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
The plot of P_T versus x₂ is a straight line. The plots of P₁ versus x₁, and P₂ versus x₂ are straight lines passing through the origin.
When x₁ = 1, x₂ = 0, P_T = \(P_{1}^{0}\) and when x₁ = 0, x₂ = 1, P_T = \(P_{2}^{0}\) as shown by lines I and II in Fig. 2.2

Question 21.
Explain the composition of vapour phase above a liquid mixture.
Answer:
Consider a liquid mixture of two liquids A and B having vapour pressures \(P_{\mathrm{A}}^{0}\) and \(P_{\mathrm{B}}^{0}\) and mole fractions x₁ and x₂ respectively in the liquid phase.
By Raoult’s law, the vapour pressures of two liquids will be,
P_A = x₁\(P_{\mathrm{A}}^{0}\) and P_B = x₂\(P_{\mathrm{B}}^{0}\)
The total vapour pressure of this liquid mixture is,
P_T = P_A + P_B
P_T = x₁\(P_{\mathrm{A}}^{0}\) + x₂\(P_{\mathrm{B}}^{0}\)
The vapour above liquid surface contains A and B. If y₁ and y₂ are the mole fractions of A and B components respectively in the vapour phase, then by Dalton’s law of partial pressures,
P_A = y₁P_T and P_B = y₂P_T
and total vapour pressure is,
P_T = y₁P_T + y₂P_T.

Question 22.
What are ideal and nonideal solutions ?
Answer:

• Ideal solutions : These are solutions which obey Raoult’s law over an entire range of concentrations at constant temperature.
• Nonideal solutions : These are solutions which do not obey Raoult’s law over the entire range of concentrations.

Question 23.
What are the characteristics of ideal solutions ?
Answer:

• The ideal solutions obey Raoult’s law over entire range of concentrations at constant temperature.
• In the formation of an ideal solution, heat is neither evolved nor absorbed and enthalpy change for mixing is zero, i.e. Δ_mix H = 0.
• In the formation of an ideal solution, there is no volume change on mixing two liquid components and the volume of solution is equal to the sum of volumes of two liquid components. Δ_mix V = 0
• In the ideal solution, solvent-solvent, solute-solute and solvent-solute interactions are comparable.
• The vapour pressure of an ideal solution lies between vapour pressures of two pure components.

Question 24.
Give an example of an ideal solution.
Answer:
A liquid mixture of benzene and toluene which have nearly identical physical properties and inter molecular forces forms an ideal solution.

Question 25.
What are the characteristics of nonideal solutions.
Answer:

• Nonideal solutions do not obey Raoult’s law over the entire range of concentrations.
• The vapour pressures of these solutions may be higher or lower than ideal solutions.
• These solutions exhibit two types of deviations from Raoult’s law namely (a) positive deviation and (b) negative deviation.
• These solutions give azeotropic mixtures.

Question 26.
Explain solutions with positive deviations from Raoult’s law.
Answer:
(i) A solution or a liquid mixture which has higher vapour pressure than theoretically calculated by Raoult’s law or higher than those of pure components is called a nonideal solution with positive deviation.

(ii) In these solutions, solute-solvent intermolecular attractions are weaker than those between solvent-solvent and solute-solute interactions.
(iii) For example, solutions of acetone and ethanol, carbon disulphide and acetone, etc.

Question 27.
Explain solutions with negative deviations from Raoult’s law.
Answer:
(1) A solution or a liquid mixture which has lower vapour pressure than theoretically calculated by Raoult’s law or lower than those of pure components is called a nonideal solution with negative deviation.

(2) In these solutions, the intermolecular interactions between solvent and solute molecules are stronger than solvent-solvent or solute-solute interactions.
(3) For example, solutions of phenol and aniline, chloroform and acetone, etc.

Solved Examples 2.5

Question 28.
Solve the following :

(1) The vapour pressures of two liquids A and B are 400 mm Hg and 600 mm Hg respectively at 47 °C. A solution is prepared by dissolving 10 g of A of molar mass 60 g mol^-1 in 80 g of B of molar mass 40 g mol^-1. Find the vapour pressure of the solution.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg.
W_A = 10 g and W_B = 80 g.
M_A = 60 g mol^-1; M_B = 40 g mol^-1, P_soln = ?
n_A = \(\frac{W_{\mathrm{A}}}{M_{\mathrm{A}}}=\frac{10}{60}\) = 0.1667 mol
n_B = \(\frac{W_{\mathrm{B}}}{M_{\mathrm{B}}}=\frac{80}{40}\) = 2 mol
Total number of moles = n = n_A + n_B
= 0.1667 + 2
= 2.1667 mol

= 0.07693 × 400 + 0.9230 × 600
= 30.772 + 553.8
= 584.572 mm Hg
Ans. Vapour pressure of solution
= P_soln = 584.572 mm Hg

(2) 2.5 mol of a liquid A is mixed with 4.5 mol of liquid B at 25 °C. If the vapour pressures of A and B at 25 °C are 160 mm Hg and 230 mm Hg respectively, calculate the vapour pressure of the liquid mixture.
Solution :
Given : n_A = 2.5 mol; n_B = 4.5 mol,
\(P_{\mathrm{A}}^{0}\) = 160 mm Hg; \(P_{\mathrm{B}}^{0}\) = 230 mm Hg, P_soln = ?
Total number of moles = n = n_A + n_B
= 2.5 + 4.5
= 7.0 mol
Mole fraction of A = x_A = \(\frac{n_{\mathrm{A}}}{n}=\frac{2.5}{7}\) = 0.3571
Mole fraction of B = 1 – x_A = 1 – 0.3571
= 0.6429
\(P_{\text {soln }}=x_{\mathrm{A}} P_{\mathrm{A}}^{0}+x_{\mathrm{B}} P_{\mathrm{B}}^{0}\)
= 0.3571 × 160 + 0.6429 × 230
= 57.136 + 147.9
= 205 mm Hg
Ans. P_soln = 205 mm Hg

(3) The vapour pressures of pure liquids A and B are 400 mm Hg and 650 mm Hg respectively at 330 K. Find the composition of liquid and vapour if total vapour pressure of solution is 600 mm Hg.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg,
P_T = 600 mm Hg, T = 330 K; x_A = ? x_B = ?, y₁ = ? y₂ = ?
(x is mole fraction in liquid phase while y is mole fraction in vapour phase.)
P_T = (\(P_{\mathrm{A}}^{0}\) – \(P_{\mathrm{B}}^{0}\))x_B + \(P_{\mathrm{A}}^{0}\)
600 = (650 – 400)x_B + 400
= 250x_B + 400
∴ x_B = \(\frac{600-400}{250}\) = 0.8
∵ x_A + x_B = 1
∴ x_A = 1 – x_B = 1 – 0.8 = 0.2
The composition of A and B in liquid mixture is, x_A = 0.2 and x_B = 0.8.
If P₁ and P₂ are vapour pressures (or partial pressures) of A and B in vapour phase then by Raoult’s law,
P₁ = x_A × \(P_{\mathrm{A}}^{0}\) = 0.2 × 400 = 80 mm Hg
P₂ = x_B × \(P_{\mathrm{B}}^{0}\) = 0.8 × 650 = 520 mm Hg
If y₁ and y₂ are mole fractions of A and B respectively in vapour phase then, by Dalton’s law,
P₁ = y₁P_T

(or y₂ = 1 – y₁ = 1 – 0.1333 = 0.8667)
Ans. Composition of liquid : x_A = 0.2 and x_B = 0.8
Composition of vapour: y_A = 0.1333 and y_B = 0.8667

(4) A mixture of two liquids A and B have vapour pressures 3.4 × 10⁴ Nm^-2 and 5.2 × 10 Nm^-2. If the mole fractions of A is 0.85, find the vapour pressure of the solution.
Solution:
Given : Vapour pressure of pure liquid A
= \(P_{\mathrm{A}}^{0}\) = 3.4 × 10⁴ Nm^-2
Vapour pressure of pure liquid B
= \(P_{\mathrm{B}}^{0}\) = 5.2 × 10⁴ Nm^-2
Mole fraction of A = x_A = 0.85
Mole fraction of B = x_B = 1 – x_A
= 1 – 0.85
= 0.15
The vapour solution is given by
\(P_{\text {soln }}=X_{A} P_{A}^{0}+X_{B} P_{B}^{0}\)
= 0.85 × 3.4 × 10 + 0.15 × 5.2 × 10⁴
=2.89 × 10⁴ + 0.78 × 10⁴
=(2.89 + 0.78) × 10⁴
P_soln = 3.67 × 10⁴ Nm^-2
Ans. Vapour pressure of a solution = 3.67 × 10⁴ Nm^-2

Question 29.
Define the term colligative property. Give examples.
Answer:
(1) Colligative Property : The property of a solution which depends on the total number of particles of the solute (molecules, ions) present in the solution and does not depend on the nature or chemical composition of solute particles is called colligative property of the solution.

(2) Examples of colligative properties : (a) lowering or relative lowering of vapour pressure of a solution (b) elevation in the boiling point (c) depression in the freezing point (d) osmotic pressure.

Question 30.
Explain and define the term vapour pressure of a liquid.
OR
What is vapour pressure of a liquid?
Answer:

(1) If a volatile liquid is placed in an open vessel, the liquid molecules have a tendency to escape in a gaseous state forming vapour and diffuse into surroundings. Hence evaporation takes place continuously, and no equilibrium is attained.

(2) If a liquid is placed in a closed vessel, the vapour molecules get accumulated on its surface. These vapour molecules are in continuous random motion. They collide with each other, with walls of a container and surface of the liquid, and return to the liquid state. This reverse phenomenon is called condensation.

(3) After some time, rates of evaporation and condensation become equal and an equilibrium is established between liquid and vapour phases. At this stage the vapour exerts a constant pressure called vapour pressure on liquid surface at constant temperature.

(4) Vapour pressure : The pressure exerted by the vapour of a liquid (or solid) when it is in equilibrium with the liquid (or solid) phase at a constant temperature is called the vapour pressure of the liquid (or solid).

(5) The vapour pressure of a liquid increases with the increase in temperature.

Question 31.
Explain the following terms :
(1) Relative vapour pressure of a solution
(2) Lowering of vapour pressure of a solution
(3) Relative lowering of vapour pressure.
Answer:
(1) Relative vapour pressure of a solution : If Po is the vapour pressure of a pure liquid (solvent) and P is the vapour pressure of a solution after adding a nonvolatile solute, then, relative vapour pressure = \(\frac{P}{P_{0}}\).

(2) Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.
If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ – P) is the lowering of the vapour pressure.

(3) Relative lowering of vapour pressure : If P₀ and P are the respective vapour pressures of a pure liquid (solvent) and the solution containing a non-volatile solute then P < P₀. Hence, P₀ – P represents the lowering of the vapour pressure due to addition of a nonvolatile solute.
∴ Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}=\frac{\Delta P}{P_{0}}\)

Question 32.
State and explain Raoult’s law for solutions of nonvolatile solutes.
Answer:
(a) Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

(b) Explanation : Let P0 and P be the vapour pressures of a pure solvent and a solution respectively. If x₁ is the mole fraction of the solvent then Raoult’s law can be represented as,
P = x₁P₀
For a binary solution containing one solute, if x₁ and x₂ are mole fractions of a solvent and a solute respectively then,
x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P = x₁P₀
= (1 – x₂) P₀
= P₀ – x₂P₀
= P₀ – P = x₂P₀
∴ P₀ – P = x₂P₀
∴ x₂ = \(\frac{P_{0}-P}{P_{0}}\)
P₀ – P = ΔP is the lowering of vapour pressure
∴ x₂ = \(\frac{\Delta P}{P_{0}}\)
In the equation, P₀ – P/P₀ is called relative lowering of vapour pressure.
Hence Raoult’s law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute.

Question 33.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
Consider a binary solution containing a nonvolatile solute. If P₀ and P are vapour pressures of a pure solvent and the solution respectively then, Lowering of vapour pressure = ΔP = P₀ – P
Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law,
\(\frac{P_{0}-P}{P_{0}}=x_{2}\)
where x₂ is mole fraction of the solute. Therefore the relative lowering of vapour pressure is a colligative property.

Question 34.
Explain the variation of vapour pressure with mole fraction of a solvent in solution.
OR
Explain the variation of vapour pressure with the concentration of a solution.
Answer:
The vapour pressure of a pure solvent decreases when a nonvolatile solute is dissolved in it. Consider a pure solvent with vapour pressure P₀ and mole fraction x₁. For a solution containing a non-volatile solute, if x₁ and x₂ are the mole fractions of a solvent and a solute respectively, then x₁ + x₂ = 1 and x₁ < 1.

By Raoult’s law, P = x₁ × P₀. As the mole fraction of a solvent in the solution increases the vapour pressure increases as shown in the above figure 2.5. When x₁ becomes equal to 1, the vapour pressure becomes P₀, i.e. the vapour pressure of a pure solvent.

If at any mole fraction of a solvent, the vapour of the solution is P, then the lowering of vapour pressure will be, ΔP = P₀ – P.

Question 35.
Derive a relation between relative lowering of vapour pressure and molar mass of non-volatile solute.
Answer:
Consider a solution in which W₁ gram of a solvent of molar mass (or molecular weight) M₂ contains W₂ gram of a solute of molar mass M₂. Then
number of moles of a solvent = n₁ = \(\frac{W_{1}}{M_{1}}\)
Number of moles of a solute = n₂ = \(\frac{W_{2}}{M_{2}}\)
∴ Total number of moles = n = n₁ + n₂
Mole fraction of the solvent = x₁ = \(\frac{n_{1}}{n}\)
Mole fraction of the solute = x₂ = \(\frac{n_{2}}{n}\)

In the case of an ideal solution which is a dilute solution, the concentration and the number of moles of the solute are very low, i.e. n₂ << n₁

If P₀ and P are the vapour pressures of a pure solvent and a solution respectively, then relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law

Hence by measuring the vapour pressure of a pure solvent and a solution, the molar mass of the dissolved nonvolatile substances can be determined.

Solved Examples 2.6 – 2.7

Question 36.
Solve the following :

(1) The vapour pressure of a pure liquid at 298 K is 4 × 10⁴ Nm^-2. When a nonvolatile solute is dissolved the vapour pressure becomes 3.65 × 10⁴ Nm^-2. Calculate (A) relative vapour pressure, (B) lowering of vapour pressure and (C) relative lowering of vapour pressure.
Solution :
Given : P₀ = 4 × 10⁴ Nm^-2
P = 3.65 × 10⁴ Nm^-2
(A) Relative vapour pressure = \(\frac{P}{P_{0}}\)
= \(\frac{3.65 \times 10^{4}}{4 \times 10^{4}}\)
= 0.9125

(B) Lowering of vapour pressure = ΔP = P₀ – P
= 4 × 10⁴ – 3.65 × 10⁴
= (4 – 3.65) × 10⁴
= 0.35 × 10⁴ Nm^-2 = 3.5 × 10³ Nm^-2

(C) Relative lowering of vapour pressure is given by,

Ans. (A) 0.9125 (B) 3.5 × 10³ Nm^-2 (C) 0.0875

(2) A pure liquid has vapour pressure 5.2 × 10⁴ Pa at 298 K. When a solute is dissolved, the mole fraction of it is 0.02 in the solution. Find the vapour pressure of the solution.
Solution :
Given : P₀ = 5.2 × 10⁴ Pa
x₂ = 0.02
P = ?

Ans. Vapour pressure of the solution
= 5.096 × 10⁴ Pa

(3) The vapour pressure of pure benzene is 640 mm of Hg. 2.175 × 10^-3 kg of nonvolatile solute is added to 39 gram of benzene, the vapour pressure of solution is 600 mm of Hg. Calculate molar mass of solute (C = 12, H = 1).
Solution :
Given : P₀ = 640 mm Hg
W₁ = 39 g benzene = 39 × 10^-3 kg
W₂ = 2.175 × 10^-3 kg
P = 600 mm Hg
M₁ = 78 × 10^-3 kg, M₂ = ?

Ans. Molar mass of a solute = 69.6 × 10^-3 kg mol^-1

(4) In an experiment, 18.04 g of mannitol were dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol.
Solution :
Given : Mass of a solute (mannitol)
= W₂ = 18.04 g
Mass of a solvent (water) = W₁ = 100 g
Vapour pressure of a solvent (water)
= P₀
= 17.535 mm Hg
Lowering of vapour pressure = ΔP = 0.309 mm Hg
Molar mass of H₂O = M₁ = 18 g mol^-1
Molar mass of solute (mannitol) = M₂ = ?

Ans. Molar mass of mannitol = 184.3 g mol^-1

(5) The vapour pressure of 2.1% solution of a non-electrolyte in water at 100 °C is 755 mm Hg. Calculate the molar mass of the solute.
Solution :
Given : At 100 °C, vapour pressure of water = P₀ = 760 mm Hg
Vapour pressure of the solution = P = 755 mm Hg
Since the solution is 2.1 % by mass,
Mass of solute (nonelectrolyte) = W₂ = 2.1 g
Mass of water = W₁ = 100 – 2.1 = 97.9 g
Molar mass of water = M₁ = 18 g mol^-1
Molar mass of nonelectrolyte = M₂ = ?

Ans. Molar mass of nonelectrolyte = 58.69 g mol^-1

(6) Calculate the mass of a nonvolatile solute (molar mass 40 × 10^-3 kg/mol) which is dissolved in 114 × 10^-3 kg octane to reduce its vapour pressure to 80%.
Solution :
Given : Molar mass of solute = M₂ =40 × 10^-3 kg mol^-1
Mass of solvent (octane) = W₁ = 114 × 10^-3 kg
If vapour pressure of octane = P₀ = 100
Vapour pressure of solution = P = 80
Molar mass of octane (C₈H₁₈) = 114 × 10^-3 kg mol^-1
Mass of solute = W₂ = ?
In this solution, since the vapour pressure is decreased by greater extent, 100 – 80 = 20% the solution must be concentrated and number of moles of solute must be large so that n₁ + n₂ ≠ n₁
Hence Raoult’s law must be represented as,

∴ Mass of solute dissolved = W
= moles × molar mass = 0.25 × 40 × 10^-3
= 0.01 kg
Ans. Mass of solute dissolved = 0.01 kg

(7) The vapour pressure of water is 16.8 mm Hg at a certain temperature. If the vapour pressure of the solution is 16.78 mm, find the molality of the solution.
Solution :
Given : P₀ = 16.8 mm Hg; P = 16.78 mm Hg
Molar mass of water = M₁ = 18 g mol^-1
Molality of the solution = m = ?

Ans. Molality of the solution = 0.66 m

Question 37.
Explain the effect of temperature on the vapour pressure of a liquid.
Answer:
The vapour pressure of a liquid is the pressure of the vapour in equilibrium with the liquid at a given temperature. The evaporation of a liquid requires thermal energy. Hence, as temperature rises, the vapour pressure rises until it becomes equal to the external pressure, generally the atmospheric pressure, 101.3 kNm^-2 (1 atm). This temperature is called the normal boiling point of the liquid.

Question 38.
What is boiling point of liquid?
OR
Define boiling point.
Answer:
The boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, i.e., the atmospheric pressure (1 atm), e.g., the boiling point of water at 1 atm is 373 K. m

Question 39.
Explain the elevation in the boiling point of a solution.
Answer:
The elevation in the boiling point of a solution is defined as the difference between the boiling points of the solution and the pure solvents at a given pressure, e.g. If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in boiling point, ΔT_b = T – T₀. It is a colligative property.

Question 40.
What are the units of molal elevation constant?
Answer:
Boiling point elevation, ΔT_b is given by,
ΔT_b = K_b × m
where m is molality in mol kg^-1 and K_b is molal elevation constant or abullioscopic constant.

∴ K_b has units K kg mol^-1 (or °C kg mol^-1)
Therefore, molal elevation constant is the elevation in boiling point produced by 1 molal solution of a nonvolatile solute.

Question 41.
Derive the relation between molar mass of the solute and boiling point elevation.
Answer:
The boiling point elevation, ΔT_b of a solution is directly proportional to molality (m) of the solution.
∴ ΔT_b ∝ m
ΔT_b = K_b m
where K_b is a proportionality constant
If m = 1 molal, then
ΔT_b = K_b
where K_b is called molal elevation constant.
The molality of the solution is given by,
Number of moles of the solute, m = \(\frac{\text { Number of moles of the solute }}{\text { Weight of the solvent in } \mathrm{kg}}\)
Let W₁ = Weight (in gram) of a solvent,
W₂ = Weight (in gram) of a solute
M₂ = Molecular weight of the solute
Then the molality (m) of the solution is given by

If the weights and molecular weight are expressed in kg, then,
\(\Delta T_{\mathrm{b}}=K_{\mathrm{b}} \times \frac{W_{2}}{W_{1} M_{2}}\)

Question 42.
Define molal elevation constant (Ebullioscopic constant)? Does it depend on the nature of a solute ? What are its units ?
Answer:
(1) Molal elevation constant (Ebullioscopic constant) : It is defined as the elevation in boiling point, produced by dissolving one mole of a solute in 1 kg (or 1000 gram) of a solvent (i.e. 1 molal solution).
The elevation in the boiling point,
ΔT_b is given by ΔT_b = K_b × m
where Kb is molal elevation constant and m is molality of the solution.
∴ When m = 1, ΔT_b = K_b
(2) K_b depends only on the nature of the solvent.
(3) K_b does not depend on the nature of the solute.
(4) It does not depend on concentration of the solution.
(5) The units of molal elevation constant are :
(A) K kg mol^-1 and (B) Km^-1.

Solved Examples 2.8

Question 43.
Solve the following :

(1) Calculate the (i) elevation in the boiling point and (ii) the boiling point of 0.05 m aqueous solution of glucose. (K_b = 0.52 Km^-1)
Solution :
Given : Concentration of the solution = m = 0.05 m
Molal elevation constant = K_b = 1.86 K kg mol^-1
Boiling point of pure water = T₀ = 373 K
Elevation in the boiling point = ΔT_b = ?
Boiling point of solution = T_b = ?

(i) ΔT_b = K_b × m
= 0.52 × 0.05
= 0.026K

(ii) The elevation in the boiling point is given by,
ΔT_b = T_b – T₀
∴ Boiling point of a solution,
T_b = T₀ + ΔT_b
= 373 + 0.026
= 373.026 K
Ans. ΔT_b = 0.026 K, T_b = 373.026 K

(2) 0.18 molal aqueous solution of a substance boils at 373.25 K. Calculate the molal elevation constant of water. (Boiling point of water is 373.15 K)
Solution :
Given : Concentration of solution = m = 0.18 m
Boiling point of water = T₀ = 373.15 K
Boiling of the solution = T_b = 373.25
Molal elevation constant = K_b = ?
Elevation in boiling point = ΔT_b
= T_b – T₀
= 373.25 – 373.15
= 0.1 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}=\frac{0.1}{0.18}\) = 0.5556 K kg mol^-1
Ans. Molal elevation constant of water
= 0.5556 K kg mol^-1

(3) The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute. [K_b for benzene = 2.53 K kg mol^-1]
Solution :
Given : \(T_{b}^{0}\) = 353.23 K; T_b = 354.11 K
W₁ = 90 g; W2 = 1.8 g; K_b = 2.53 K kg mol^-1
Molar mass = M₂ = ?
ΔT_b = T_b – \(T_{b}^{0}\) = 354.11 – 353.23 = 0.88 K

Ans. Molar mass = M₂ = 57.5 g mol^-1

(4) Boiling point of a solvent is 80.2°C. When 0.419 g of the solute of molar mass 252.4 g mol^-1 is dissolved in 75 g of the above solvent, the boiling point of the solution is found to be 80.256 °C. Find the molal elevation constant.
Solution :
Given : T₀ = (273 + 80.2) K
T_b = 273 + 80.256 (K)
W₁ = 75g
W₂ = 0.419 g
M₂ = 252.4 g mol^-1
K_b = ?
ΔT_b = T_b – T₀
= (273 + 80.256) – (273 + 80.2)
= 0.056 K

Ans. K_b = 2.53 K kg mol^-1

(5) 3.795 g of sulphur is dissolved in 100 g of CS₂. This solution boils at 319.81 K. What is the molecular formula of sulphur in solution? Theboiling point of CS₂ is 319.45 K.
(Given that K_b for CS₂ =2.42 K kg mol^-1 and atomic mass of S = 32.)
Solution :
Given : W₂ = 3.795 g; W₁ = 100 g CS₂
T_b = 319.81 K; T₀ = 319.45 K
K_b = 2.42 K kg mol^-1 M₂ = ?
ΔT_b = T_b – T₀ = 319.81 – 319.45 = 0.36 K

Number of S atoms in molecule = \(\frac{255.1}{32}\)
= 7.972
≅ 8
Ans. Formula of sulphur in CS₂ solution = S₈

(6) Calculate the mass in grams of an impurity of molar mass 100 g mol^-1 which would be required to raise the boiling point of 50 g of chloroform by 0.30°C. (K_b for chloroform = 3.63 K kg mol^-1)
Solution :
Given : M₂ = 100 g mol^-1
W₁ = 50 g chloroform
W₂ = ?
ΔT_b = 0.30°C.
K_b = 3.63 K kg mol^-1

Ans. Weight of impurity = 0.4132 g

(7) A solution containing 0.5126 g of naphthalene (molar mass = 128.17 g mol^-1) in 50.0 g of CCl₄ gives a boiling point elevation of 0.402 °C. While a solution of 0.6216 g of unknown solute in the same mass of the solvent gives a boiling point elevation of 0.647 °C. Find the molar mass of the unknown solute. (K_b for CCl₄ = 5.03 K kg mol^-1 of solvent)
Solution:
Given : For naphthalene solution :
W₂ = 0.5126 g
W₁ = 50 g
M₂ = 128.17 g mol^-1
K_b = 5.03 K kg mol^-1
ΔT_b = 0.402°C
For unknown solute :
W’₁ = 50g
W’₂ = 0.6216 g
ΔT’_b = 0.647°C
M’₂ = ?
For the solution of unknown substance,

Ans. Molecular weight of unknown substance = 96.65 g mol^-1

(8) A solution containing 0.73 g of camphor (molar mass 152 g mol^-1) in 36.8 g of acetone (boiling point 56.3°C) boils at 56.55°C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46°C. Calculate the molar mass of the unknown compound.
Solution:
Given : Mass of solute = W₂ = 0.73 g camphor
Mass of solvent = W₁ = 36.8 g
Molar Mass of solute = M₂ = 152 g mol^-1
Boiling point of acetone = T₀ = (273 + 56.3) K
Mass of unknown solute = W’₂ = 0.564 g
Mass of solvent = W’₁ = 36.8 g
Boiling point of solution of camphor = T_b = (273 + 56.55) K
Boiling point solution of unknown compound = T’_b = (273 + 56.46) K
Molar mass of unknown compound = M’₂ = ?

For camphor solution,
∴ ΔT_b = T_b – T₀
= (273 + 56.55) – (273 + 56.3)
= 0.25 K

For a solution of unknown compound
ΔT’_b = ΔT’_b – T₀
= (273 + 56.46) – (273 + 56.3)
= 0.16 K

Ans. Molar mass of the compound = 183.5 g mol^-1

(9) 0.12 molal solution of a substance boils at 373.21°K. Calculate molal elevation constant of the solvent.
Solution :
Given : Concentration of solution = m = 0.12
Boiling point of the solution = T_b = 373.21 K
For solvent (water) T₀ = 373.15 K
Molal elevation constant, K_b = ?
ΔT_b = T_b – T₀
= 373.21 – 373.15
= 0.06 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}\)
= \(\frac{0.06}{0.12}\)
=0.5 K kg mol^-1
Ans. Molal elevation constant = K_b = 0.5 K kg mol^-1

(10) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 5.02 K kg mol^-1)
Solution :
Given : Boiling point of water = T₀ = 273 + 99.63
= 372.63 K
Boiling point of a solution = T_b = 273 + 100 = 373 K
Mass of a solvent (water) = W₁ = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁) = M2
= 342 g mol^-1
K_b = 5.02 K kg mol^-1
Mass of solute (sucrose) = W₂ = ?
ΔT_b = T_b – T₀
= 373 – 372.63
= 0.37 K

Ans. Mass of sucrose required to be added = 12.60 g

(11) A solution of phosphorus prepared by dissolving 0.0175 kg of phosphorus in 0.08 kg of CS₂ has a boiling point 319.87 K. If K_b for CS₂ is 2.4 K kg mol-1 and atomic mass of phosphorus is 31 × 10^-3 kg mol^-1, find the formula of phos-phorus. (Boiling point of CS₂ = 319.45 K)
Solution :
Given : Mass of solvent (CS₂) = W₁ = 0.08 kg
Mass of phosphorus = W₂ = 0.0175 kg
Boiling point of CS₂ = T₀ = 319.45 K
Boiling point of solution = T_b = 319.87 K
Molal elevation constant = K_b = 2.4 K kg mol^-1
Atomic mass of phosphorus = 31 × 10^-3 kg mol^-1
Molecular formula of phosphorus = ?
ΔT_b = T_b – T₀ = 319.87 – 319.45 = 0.42 K

∴ Number of atoms in a molecule of phosphorus = \(\frac{\text { molar mass of phosphorus }}{\text { atomic mass of phosphorus }}\)
= \(\frac {125}{31}\)
= 4.032
≅ 4
Hence the molecular formula of phosphorus is P₄ in CS₂.
Ans. Molecular formula of phosphorus = P₄

(12) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 0.52 K kg mol^-1)
Solution :
Given : Pressure = P = 750 mm Hg
T₀ = 273 + 99.63 = 372.63 K
T_b = 273 + 100 = 373 K
K_b = 0.52 K kg mol^-1
Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 × 10^-3 kg mol^-1
W₁ = 500 g = 0.5 kg
Mass of sucrose to be added = W₂ = ?
ΔT_b = T_b – T₀ = 373 – 372.63 = 0.37 K

Ans. Mass of sucrose to be added = 121.7 × 10^-3 kg
= 121.7 g

(13) 35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.
Solution :
35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O.
Molar mass of water = 18 g mol^-1
Molar mass of ethylene glycol (CH₂OH-CH₂OH) = 62 g mol^-1
Number of moles of water = n₁ = \(\frac{65}{18}\) = 3.611 mol
Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)
= 0.5645 mol
Total moles = n = n₁ + n₂ = 3.611 + 0.5645
= 4.1755 mol
Mole fraction of ethylene glycol = x₂
\(=\frac{n_{2}}{n_{1}}=\frac{0.5645}{4.1755}\) = 0.1352
∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352
= 0.8648
Ans. Mole fraction of water = 0.8648
Mole fraction of ethylene glycol = 0.1352

Question 44.
Define freezing point of a liquid.
Answer:
The freezing point of a liquid is defined as the temperature at which the solid coexists in the equilibrium with the liquid and the vapour pressure of the liquid and the solid are equal.

Question 45.
Explain the depression in the freezing point of a solution.
Answer:
The depression in the freezing point of a solution is defined as the difference between the freezing points of a pure solvent and that of the solution.
If T₀ and T are the respective freezing points of a pure solvent and a solution, then the depression in the freezing point ΔT_f is given by,
ΔT_f = T₀ – T (T < T₀)
The depression in the freezing point (ΔT_f) is a colligative property.

Question 46.
What causes depression in freezing point ?
OR
Explain, freezing point depression as a consequence of vapour pressure lowering.
Answer:
The freezing point of a liquid is the temperature at which the liquid and the solid have the same vapour pressure.

Addition of a nonvolatile solute to a liquid decreases the freezing point, i.e., the freezing point of the solution is less than that of the pure solvent. This is due to the lowering of the vapour pressure of the solvent by the addition of the nonvolatile solute.

When a liquid is cooled from the point A, its vapour pressure decreases and at the point B, it freezes (solidifies).

In case of a solution, since the vapour pressure is lowered, the freezing point decreases. Hence, if the solution is cooled from the point A’, it freezes at lower temperature B’, than the pure liquid. This is also due to separation of solvent molecules due to solute molecules decreasing their intermolecular attraction.

If T₀ and T are the freezing points of a pure solvent and the solution, then, the depression in the freezing point is given by,
ΔT_f = T₀ – T.
This depression ΔT_f depends on the lowering of the vapour pressure (P₀ – P). ΔT_f ∝ (P₀ – P), where P₀ and P are the vapour pressures of the pure liquid and the solution.

Question 47.
What is a relationship between freezing point depression and concentration of solute (or solution)?
Answer:
It is observed experimentally that as the concentration of a solution increases, the freezing point of the solution decreases and hence the depression in the freezing point (ΔT_f) increases.

The depression in the freezing point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Thus,
ΔT_f ∝ m
where m is the molality of the solution.
∴ ΔT_f = Kfm, where Kf is a constant of proportionality. If m = 1 molal,
ΔT_f = K_f. Hence K_f is called the cryoscopic constant or molal depression constant. K_f is characteristic of the solvent.

Question 48.
What are the units of molal depression constant or cryoscopic constant?
Answer:
The freezing point depression, ΔT_f is given by,
ΔT_f = K_f × m
where m is molality in mol kg^-1 and K_f is molal depression constant or cryoscopic constant.

∴ K_f has unit K kg mol^-1 (or °C kg mol^-1)
Therefore cryoscopic constant is the depression in freezing point produced by 1 molal solution of a nonvolatile solute.

Question 49.
Write the formula to determine molar mass of a solute using freezing point depression method.
Answer:
M₂ = \(\frac{K_{\mathrm{f}} \times W_{2} \times 1000}{W_{1} \times \Delta T_{\mathrm{f}}}\)
where
K_f = Molal depression constant
ΔT_f = Depression in freezing point
W₁ = Mass of a solvent
W₂ = Mass of a solute.
M₂ = Molar mass of solute

Question 50.
Define cryoscopic constant (or molal depression constant).
Answer:
Molal depression constant : It is defined as the depression in freezing point, produced by dissolving one mole of a solute in 1 kg (or 1000 g) of a solvent (i.e. 1 molal solution).

Solved Examples 2.9

Question 51.
Solve the following :

(1) 1.35 g of a substance when dissolved in 55 g acetic acid produced a depression of 0.618°C in a freezing point. Calculate the molar mass of the dissolved substance. (K_f = 3.865 K kg mol^-1)
Solution :
Given : Mass of a solvent = W₁ = 55 g
Mass of a solute = W₂ = 1.35 g
Depression in freezing point = ΔT_f = 0.618 °C
Molal depression constant = K_f = 3.865 K kg mol^-1
Molar mass of a solute = M₂ = ?

Ans. Molar mass of the substance = 153.5 g mol^-1

(2) When certain amount of sucrose is dissolved in 1 kg of water, the freezing point of the solution is found to be 272.8 K. If the molecular mass of sucrose is 342 g mol^-1 and K_f for water is 1.86 K kg mol^-1, calculate the amount of sucrose present in the solution.
Solution :
Given : W₁ = Mass of the solvent = 1 kg
T₀ = Freezing point of pure water = 273 K
T_f = Freezing point of the solution = 272.8 K
ΔT_f = T₀ – T_f = Depression in freezing point
= 273 K – 272.8 K
= 0.2 K
M₂ = Molecular mass of the solute
= 342 × 10^-3 kg mol^-1
K_f = Molal freezing point depression constant for water
= 1.86 K kg mol^-1
W₂ = Mass of the solute = ?
ΔT_f = T₀ – T_f

Ans. Mass of sucrose = 36.78 × 10^-3 kg

(3) The freezing point of a pure solvent is 315 K. On addition of 0.5 mole of urea in 1 kg of the solvent, the freezing point decreases by 3 K. Calculate the molal depression constant for the solvent.
Solution :
Given :
m = Molality of urea = 0.5 m
ΔT_f = Depression in the freezing point = 3 K
K_f = Molal depression constant for the solvent = ?
ΔT_f = K_f × m
∴ K_f = \(\frac{\Delta T_{\mathrm{f}}}{m}\)
= \(\frac{3}{0.5}\)
= 6 K kg mol^-1
Ans. Molal depression constant = 6 K kg mol^-1

(4) The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2g of a solute having molecular weight 100 is added to 100 g of benzene. (K_f for benzene = 5.12 K kg mol^-1)
Solution :
Given : K_f for benzene = 5.12 K kg mol^-1
T₀ = Freezing point of the solvent = 278.4 K
W₁ = Mass of the solvent = 100 g = 0.1 kg
W₂ = Mass of the solute = 2g = 2 × 10^-3 kg
M₂ = Molecular mass of the solute = 100 g mol^-1
= 100 × 10^-3 kg mol^-1
T = Freezing point of the solution = ?

Ans. Freezing point of the solution = 277.376 K

(5) 0.635 × 10^-3 kg of a substance of molar mass 190 × 10^-3 kg mol^-1 was dissolved in 30.5 × 10^-3 kg of a solvent. If the depression in the freezing point is 0.62 °C, find the molal depression constant of the solvent.
Solution :
Given : Mass of solvent = W₁ = 30.5 × 10^-3 kg
Mass of solute = W₂ = 0.635 × 10^-3 kg
Depression in freezing point = ΔT_f = 0.62 °C (or K)
Molar mass of the substance = M₂ = 190 × 10^-3 kg mol^-1
Molal depression constant = K_f = ?

Ans. Molal depression constant = K_f
= 5.66 K kg mol^-1

(6) The boiling point of an aqueous solution is 100.18 °C. Find the freezing point of the solution. (Given : K_b = 0.52 K kg mol^-1, K_f = 1.86 K kg mol^-1)
Solution :
Given : T_b = 100.18 °C + 273 = 373.18 K
K_b = 0.52 K kg mol^-1; K_f = 1.86 K kg mol^-1
Boiling point of water = T₀ = 373 K
T_f = ?
ΔT_b = T_b – T₀ = 373.18 – 373 = 0.18 K
If m is the molality of the solution, then
ΔT_b = K_b × m and ΔT_f = K_f × m

Freezing point of water = T₀ = 273 K
ΔT_f = T₀ – T_f
Hence the freezing point of the solution is,
T_f = T₀ – ΔT_f = 273 – 0.6438 = 272.3562 K
OR Freezing point of solution is -0.6438 °C.
Ans. Freezing point of the solution
= 272.3562 K
= -0.6438 °C

(7) 1.0 × 10^-3 kg of urea when dissolved in 0.0985 kg of a solvent, decreases freezing point of the solvent by 0.211 K. 1.6 × 10^-3 kg of another nonelectrolyte solute when dissolved in 0.086 kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)
Solution :
Given : Mass of Urea = W₂ = 1 × 10^-3 kg
Mass of solvent = W₁ = 0.0985 kg
Depression in freezing point = ΔT_f = 0.211 K
Molar mass of urea = M₂ = 60 g mol^-1
= 60 × 10^-3 kg mol^-1
Mass of another solute = W’₂ = 1.6 × 10^-3 kg
Mass of solvent = W’₁ = 0.086 kg
Depression in freezing point = ΔT’_f = 0.34 K
Molar mass of another solute = 60 × 10^-3 kg mol^-1

From urea solution:


For solution of another solute :

Ans. Molar mass of solute = 68.23 g mol^-1.

Question 52.
What do you understand by the terms :
(1) permeable membrane
(2) semipermeable membrane?
Answer:
(1) Permeable membrane : A membrane which allows free transfer of the solute molecules from a solution of a higher concentration to a solution of a lower concentration through it is called a permeable membrane and the transfer is called diffusion, e.g., a membrane of a paper.

(2) Semipermeable membrane : A membrane which allows free passage of only the solvent molecules but not the large solute molecules or ions of large molecular mass from a solution of a lower concentration (or a pure solvent) to a solution of higher concentration through it, is called a semi-permeable membrane, e.g., parchment paper, complex like Cu₂[Fe(CN)₆], etc.

Question 53.
Define and explain osmosis.
Answer:
(1) Definition : It is defined as a spontaneous uni-directional flow of the solvent molecules from a pure solvent or a dilute solution to the more concentrated solution through a semipermeable membrane.

Example : A flow of water molecules from a dilute solution into a concentrated glucose solution through a parchment paper.

(2) Explanation : Consider a vessel divided into two compartments by a semipermeable membrane. When one compartment is filled with a pure solvent or a dilute solution and another by concentrated solution, there is a spontaneous of solvent molecules to the concentrated solution. This arises due to higher vapour pressure of a pure solvent or dilute solution than concentrated solution.

Question 54.
Explain the relation between osmotic pressure and concentration of solution.
Ans.
(1) Consider V dm³ of a solution in which n₁ moles of a solvent contains n₂ moles of a nonvolatile solute at absolute temperature T.

(2) The osmotic pressure, n of a solution is given by,
π = \(\frac{n R T}{V}\)
R is gas constant having value 0.08206 dm³ atm K^-1 mol^-1 (OR L atm K^-1 mol^-1). Since concentration, C of a solution is in mol dm^-3 or molarity is,
C = \(\frac{n}{V}\) mol dm^-3 or M
∴ π = CRT
(If concentration C is expressed in mol m^-3 and R = 8.314 J K^-1mol^-1, then π will be in SI units, pascals or Nm^-2.)

Question 55.
Explain the terms :
(1) Isotonic solutions
(2) Hypotonic solutions
(3) Hypertonic solutions.
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A = \(\frac{n_{\mathrm{A}}}{V}\) (in mol dm^-3) and
C_B = \(\frac{n_{\mathrm{B}}}{V}\) (in mol dm^-3)
If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A = C_ARTand π_B = C_BRT where π_A and π_B are their osmotic pressures.
For the isotonic solutions,
π_A = π_B
∴ C_A = C_B
∴ \(\frac{n_{\mathrm{A}}}{V}=\frac{n_{\mathrm{B}}}{V}\)
∴ n_A = n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypotonic solutions : When two solutions have different osmotic pressures, then the solution having lower osmotic pressure is said to be a hypotonic solution with respect to the other solution.
Explanation : Consider two solutions of the substances A and B having osmotic pressures π_A and π_B. If π_B is less than π_A, then the solution B is a hypotonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then, C_B < C_A. Hence, for equal volumes of the solutions, n_A < n_A.

(3) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.
Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B. If π_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then C_B > C_A. Hence, for equal volume of the solutions, n_B > n_A.

Question 56.
Explain colligative properties of electrolytes.
Answer:

• The electrolytic solutions do not exhibit colligative properties similar to nonelectrolytes.
• The colligative properties of electrolytes are higher than those shown by equimolar solutions of nonelectrolytes.
• The molar masses of electrolytes determined by colligative properties are found to be considerably lower than their actual molar masses.

Question 57.
Why are the colligative properties of electrolytic solutions greater than those for non-electrolytic solutions with same concentration ?
Answer:

• The electrolytes in polar solvents or aqueous solutions dissociate into two or more ions whereas nonelectrolytes do not dissociate.
• Consequently the number of particles in electrolytic solutions are considerably higher than equimolar nonelectrolytic solutions.
• Therefore the colligative properties of electrolytes are greater than nonelectrolytes with same concentration in solution.

Question 58.
What is abnormal colligative property? Explain the reasons.
Answer:
Abnormal colligative property : When the experimentally measured colligative property of a solution is different from that calculated theoretically by the van’t Hoff equation or by the laws of osmosis, then the solution is said to have abnormal colligative property.

Explanation : The colligative property depends on the number of solute particles in the solution but it is independent of their nature. Abnormal values of them arise when the dissolved solute undergoes a molecular change like dissociation or association in the solution.
The observed colligative property (or abnormal colligative property ) may be higher or lower than the theoretical value.

(i) Dissociation of the solute molecules : When a solute like an electrolyte is dissolved in a polar solvent like water, it undergoes dissociation, which results in the increase in the number of particles in the solution.

Hence, the observed value of the colligative property becomes higher than the theoretical value, e.g., when one mole of KCl is dissolved in the solution then due to dissociation, KCl → K⁺ + Cl^–, the number of particles increases, hence, the colligative properties like osmotic pressure elevation in the boiling point, etc. increase.

(ii) Association of the solute molecules : When a solute like a nonelectrolyte is dissolved in a nonpolar solvent like benzene, it undergoes association forming molecules of higher molecular mass. Hence, the number of the particles in the solution decreases. Therefore the colligative properties like osmotic pressure, elevation in the boiling point, etc., are lower than the theoretical value, e.g., nA → An.
2CH₃COOH → (CH₃COOH)₂
2C₆H₅COOH → (C₆H₅COOH)₂

Question 59.
Explain abnormal osmotic pressure.
Answer:

• When the experimentally observed osmotic pressure is different than theoretically calculated value by van’t Hoffs equation then it is called abnormal osmotic pressure.
• This arises when the dissolved solute undergoes a molecular change like association or dissociation.

Question 60.
Explain abnormal molecular masses.
Answer:
When the observed molecular masses obtained from their colligative properties of the substances are different (higher or lower) than the theoretical or normal values calculated from their molecular formulae, then they are called abnormal molecular masses.

Question 61.
How is van’t Hoff factor related to molecular mass of the substance ?
Answer:
The van’t Hoff factor is also defined as,
actual moles of particles in solution after


In case of nonelectrolytes, i < 1.
In case of electrolytes, i > 1. For example for KNO₃ and NaCl, i = 2, for Na₂SO₄, CaCl₂, i = 3, etc.

Question 62.
Write modification of expressions of colligative properties with the help of van’t Hoff factor.
Answer:
The modified expressions of colligative properties with the help of van’t Hoff factor i are as follows :
(1) By Raoult’s law :

Question 63.
Obtain a relation between degree of dissociation and molar mass for an electrolyte.
Answer:
Consider an electrolyte AxBy.

If i is van’t Hoff factor and n is total number of ions produced from dissociation of one mole of electrolyte then, the degree of dissociation α is given by,
α = \(\frac{i-1}{n-1}\)
Now if M_th and M_ob are theoretical arid observed molecular (or molar) masses respectively, then,
i = \(\frac{M_{\mathrm{th}}}{M_{\mathrm{ob}}}\)
∵ α(n – 1) = i – 1

This is a relation between degree of dissociation and molecular masses of the dissolved electrolyte AxBy.

Question 64.
Solve the following :

(1) 300 mL solution at 27 °C contains 0.2 mol of a nonvolatile solute. Calculate osmotic pressure of the solution.
Solution :
Given : V= 300 ml = 0.3 dm³ ;
T= 273 + 27 = 300 K
π = 0.2 mol, π = ?
π = \(\frac{n R T}{V}\)
= \(\frac{0.2 \times 0.08206 \times 300}{0.3}\)
= 16.41 atm
Ans. Osmotic pressure = π = 16.41 atm.

(2) 200 mL solution at 27 °C contains 10 g of a nonvolatile solute of molar mass 65 g mol^-1. What is osmotic pressure of the solution ?
Solution :
Given : V = 200 mL = 0.2 dm³; W = 10 g
M = 65 g mol^-1; T= 273 + 27 = 300 K; π = ?

Ans. Osmotic pressure = π = 18.93 atm.

(3) Calculate the osmotic pressure of 0.2 M glucose solution at 300 K. (R = 8.314 J mol^-1 K^-1)
Solution :
Given : Concentration of the solution
= C = 0.2 M
= 0.2 mol dm^-3
= 0.2 × 10³ mol m^-3
Temperature = T= 300 K
= 8.314 J mol^-1 K^-1
The osmotic pressure, π is given by,
π = CRT
= 0.2 × 10³ × 8.314 × 300
= 4.988 × 10⁵ Nm^-2 (or Pa)
Ans. Osmotic Pressure = 4.988 × 10⁵ Nm^-2

(4) A solution of cane sugar containing 18 g L^-1 has an osmotic pressure 1.25 atm. Calculate the temperature of the solution. (Molar mass of cane sugar = 342, R = 0.082 lit atm mol^-1 K^-1)
Solution:
Given : Amount of cane sugar = W = 18 g L^-1
Osmotic pressure = π = 1.25 atm
Molar mass of cane sugar = M = 342 g mol^-1
Temperature = T = ?
Number of moles of cane sugar
= \(\frac{W}{M}\)
= \(\frac{18}{342}\)
= 0.05263 mol
∴ Concentration of solution = C
= \(\frac{n}{V}\)
= \(\frac{0.05263}{1}\)
= 0.05263 mol lit^-1
π = CRT
∴ T = \(\frac{\pi}{\mathcal{C R}}\)
= \(\frac{1.25}{0.05263 \times 0.08206}\)
= 289.4 K
Ans. Temperature of solution = 289.4 K

(5) Equal volumes of two solutions, one containing glucose and another urea have osmotic pressures 12.6 atm and 23.8 atm at 25 °C. Calculate the ratio of number of moles of urea to glucose.
Solution :
Given : Osmotic pressure of glucose solution = π₁
= 12.6 atm
Osmotic pressure of urea solution = π₂
= 23.8 atm
Number of moles of glucose = n₁
Number of moles of urea = n₂

(6) One litre of a solution containing 25 g glucose has osmotic pressure 3.4 atm at 300 K. If by diluting the solution the osmotic pressure becomes 2.0 atm at the same temperature, what would be its concentration ? (Molar mass of glucose = 180)
Solution :
Given : Mass of solute (glucose) = 25 g
Volume of solution = 1.0 L
Osmotic pressure = π₁ = 3.4 atm
After dilution osmotic pressure = π₂ = 2 atm
Final concentration = c₂ = ?
The number of moles of glucose = n = \(\frac{25}{180}\)
= 0.1389 mol
Initial concentration = c₁ = \(\frac{n}{V}\)
= \(\frac{0.1389}{1}\)
At constant temperature, by van’t Hoff-Boyle’s law,

Ans. Concentration of solution = 0.0817 mol L^-1

(7) A solution of a substance having mass 1.8 × 10^-3 kg has the osmotic pressure of 0.52 atm at 280 K. Calculate the molar mass of the substance used.
[Volume = 1 dm³, R = 8.314 JK^-1 mol^-1]
Solution :
Given : W= 1.8 × 10^-3 kg = 1.8 g
π = 0.52 atm; V = 1 dm³
T = 280 K;
Molar mass = M = ?

Ans. Molar mass = M = 79.53 g mol^-1

(8) An organic substance (M = 169 gram mol^-1) is dissolved in 2000 cm³ of water. Its osmotic pressure at 12 °C was found to be 0.54 atm. If R = 0.0821 L atm K^-1 mol^-1, calculate the mass of the solute.
Solution :
Given : M = 169 g mol^-1
V = 200 cm3 = 0.2 dm³
T = 273 + 125 = 285 K
π = 0.54 atm
R = 0.0821 L atm K^-1 mol^-1; W = ?

Ans. Mass of solute = 0.78 g

(9) A solution containing 10 g glucose has osmotic pressure 3.84 atm. If 10 g more glucose is added to the same solution, what will be its osmotic pressure. (Temperature remains constant)
Solution :
Given : Mass of glucose = W₁ = 10 g
Mass of more glucose added = W₂ = 10 g
Initial osmotic pressure = π₁ = 3.84 atm
Final osmotic pressure = π₂ = ?
Total mass of glucose in final solution = W₁ + W₂
W’ = 10 + 10
= 20 g

∴ π₂ = \(\pi_{1} \times \frac{W^{\prime}}{W_{1}}\)
= \(3.84 \times \frac{20}{10}\) = 7.68 atm
Ans. Osmotic pressure of the solution = 7.68 atm

(10) A 0.1 m solution of K₂SO₄ in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_f for water is 1.86 K kg mol^-1?
Solution :
Given : m = 0.1 m, ΔT_f = 0 – (-0.43) = 0.43 °C
K_f = 1.86 K kg mol^-1, i = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{0.43}{1.86 \times 0.1}=2.312\)
Ans. van’t Hoff factor = i = 2.312

Multiple Choice Questions

Question 65.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. A molal solution is one that contains one mole of solute in
(a) one litre of the solvent
(b) 1000 g of the solvent
(c) one litre of the solution
(d) 22.4 litre of solution
Answer:
(b) 1000 g of the solvent

2. 10.0 grams of caustic soda when dissolved in 250 cm³ of water, the resultant molarity of solution is
(a) 0.25 M
(b) 0.5 M
(c) 1.0 M
(d) 0.1 M
Answer:
(c) 1.0 M

3. 5 × 10-3 kg of urea is dissolved in 2 × 10^-2 kg of water. The percentage by weight of urea is
(a) 15%
(b) 20%
(c) 25%
(d) 30%
Answer:
(b) 20%

4. Vapour pressure of solution of a nonvolatile solute is always
(a) equal to the vapour pressure of pure solvent
(b) higher than vapour pressure of pure solvent
(c) lower than vapour pressure of pure solvent
(d) constant
Answer:
(c) lower than vapour pressure of pure solvent

5. According to the Raoult’s law, the relative lowering of vapour pressure is equal to the
(a) mole fraction of solvent
(b) mole fraction of solute
(c) independent of mole fraction of solute
(d) molality of solution
Answer:
(b) mole fraction of solute

6. Partial pressure of solvent in solution of non-volatile solute is given by equation,
(a) P = x₂P⁰
(b) P⁰ = xP
(c) P = x₁P⁰
(d) P⁰ = x₁P
Answer:
(c) P = x₁P⁰

7. When partial pressure of solvent in solution of nonvolatile solute is plotted against its mole fraction, nature of graph is
(a) a straight line passing through origin
(b) a straight line parallel to mole fraction of solvent
(c) a straight line parallel to vapour pressure of solvent
(d) a straight line intersecting vapour pressure axis
Answer:
(a) a straight line passing through origin

8. Colligative property depends only on ………………. in a solution.
(a) Number of solute particles
(b) Number of solvent particles
(c) Nature of solute particles
(d) Nature of solvent particles
Answer:
(a) Number of solute particles

9. Lowering of vapour pressure of solution
(a) is a property of solute
(b) is a property of solute as well as solvent
(c) is a property of solvent
(d) is a colligative property
Answer:
(d) is a colligative property

10. Raoult’s law is not applicable to
(a) solutions of volatile solutes
(b) solutions of electrolytes
(c) dilute solutions
(d) concentrated solutions
Answer:
(d) concentrated solutions

11. The relative lowering of vapour pressure of a solution is proportional to the
(a) mole fraction of the solvent
(b) mole fraction of the solute
(c) amount of the substance
(d) volume of the solvent
Answer:
(b) mole fraction of the solute

12. The vapour pressure of an aqueous solution of glucose at 100 °C is 710 mm Hg. Hence the molality of the solution is
(a) 2.83 m
(b) 3.65 m
(c) 16.47 m
(d) 12.5 m
Answer:
(b) 3.65 m

13. Relative vapour pressure lowering depends only on
(a) Mole fraction of solute
(b) Nature of solvent
(c) Nature of solute
(d) Nature of solute and solvent
Answer:
(a) Mole fraction of solute

14. A solution having mole fraction of a solute equal to 0.05 has vapour pressure 20 × 10³ Nm^-2. Hence the vapour pressure of a pure solvent is
(a) 21.05 × 10³ Nm^-2
(b) 4 × 10⁵ Nm^-2
(c) 1 × 10³ Nm^-2
(d) 2 × 10⁴ Nm^-2
Answer:
(a) 21.05 × 10³ Nm^-2

15. The addition of the nonvolatile solute into the pure solvent ……………..
(a) increases the vapour pressure of solvent
(b) decreases the boiling point of solvent
(c) decreases the freezing point of solvent
(d) increases the freezing point of solvent
Answer:
(c) decreases the freezing point of solvent

16. A solution having the highest vapour pressure is
(a) 1 M Al₂(SO₄)₃
(b) 0.1 M NaNO₃
(c) 1 M BaCl₂
(d) 1 M Ca(NO₃)₂
Answer:
(b) 0.1 M NaNO₃

17. Molal elevation constant is elevation in boiling point produced by
(a) 1 g of solute in 100 g of solvent
(b) 100 g of solute in 1000 g of solvent
(c) 1 mole of solute in one litre of solvent
(d) 1 mole of solute in one kg of solvent
Answer:
(d) 1 mole of solute in one kg of solvent

18. The determination of molar mass from elevation in boiling point is called
(a) cryoscopy
(b) osmometry
(c) ebullioscopy
(d) spectroscopy
Answer:
(c) ebullioscopy

19. Which of the following aqueous solutions will have minimum elevation in boiling point ?
(a) 0.1 M KCl
(b) 0.05 M NaCl
(c) 1 M AIPO₄
(d) 0.1 M MgSO₄
Answer:
(b) 0.05 M NaCl

20. Which of the following solutions shows maximum depression in freezing point ?
(a) 0.5 M Li₂SO₄
(b) 1 M NaCl
(c) 0.5 M Al₂ (SO₄)₃
(d) 0.5 M BaCl₂
Answer:
(c) 0.5 M Al₂ (SO₄)₃

21. If the freezing point of 0.1 m solution is 272.814 K, then the freezing point of 0.2 m solution will be
(a) 545.628 K
(b) 265.64 K
(c) 272.628 K
(d) 0.482 K
Answer:
(c) 272.628 K

22. At a freezing point,
(a) Vapour pressure of a solution = Vapour pressure of a solid
(b) Vapour pressure of a liquid = Vapour pressure of a solid
(c) Vapour pressure of a liquid > Vapour pressure of a solid
(d) Vapour pressure of a solid > Vapour pressure of a liquid
Answer:
(b) Vapour pressure of a liquid = Vapour pressure of a solid

23. In osmosis
(a) solvent molecules pass from high concentration of solute to low concentration
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute
(c) solute molecules pass from low concentration to high concentration
(d) solute molecules pass from high concentration to low concentration
Answer:
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute

24. As temperature increases
(a) Osmotic pressure and vapour pressure decrease
(b) Vapour pressure and osmotic pressure increase
(c) Vapour pressure increases but osmotic pressure decreases
(d) Osmotic pressure increases but vapour pressure decreases
Answer:
(b) Vapour pressure and osmotic pressure increase

25. A mango kept in a salt solution shrinks. Hence the liquid content in mango with respect to the salt solution is
(a) isotonic
(b) hypertonic
(c) hypotonic
(d) equimolar
Answer:
(c) hypotonic

26. The osmotic pressure of 5% glucose solution at 300 K is
(a) 6.93 × 10⁵ Nm^-2
(b) 6.93 × 10² Nm^-2
(c) 6.83 Pa
(d) 6.00 × 10³ Pa
Answer:
(a) 6.93 × 10⁵ Nm^-2

27. 0.1 M solution of A has osmotic pressure xNm^-2 at 300 K. If 200 ml of A and 100 ml of 0.2 M solution of nonreactive solute B are mixed then the osmotic pressure will be
(a) 3x
(b) 0.05 x
(c) 1.33 x
(d) 0.75 x
Answer:
(c) 1.33 x

28. The solution A is twice hypertonic to the solution B at a given temperature. If the solution A contains 8.6 × 10²² molecules, then the number of molecules present in B are,
(a) 8.6 × 10²²
(b) 1.73 × 10²³
(c) 3.24 × 10²²
(d) 4.3 × 10²²
Answer:
(d) 4.3 × 10²²

29. Isotonic solutions are the solutions having the same
(a) surface tension
(b) vapour pressure
(c) osmotic pressure
(d) viscosity
Answer:
(c) osmotic pressure

30. If Kb for water is 0.52 Km^-1, the boiling point of 0.2 m solution of a nonvolatile solute will be
(a) 371.96 K
(b) 373.104 K
(c) 373.52 K
(d) 374.0 K
Answer:
(b) 373.104 K

31. The vapour pressure of water is 15.5 mm at 20 °C. The lowering of vapour pressure of 0.02 m K Br solution will be
(a) 0.0112 mm Hg
(b) 0.0056 mm Hg
(c) 0.056 mm Hg
(d) 0.31 mm Hg
Answer:
(a) 0.0112 mm Hg

32. Which of the following 0.1 M aqueous solutions will exert highest osmotic pressure ?
(a) Al₂(SO₄)₃
(b) Na₂SO₄
(c) MgCl₂
(d) KCl
Answer:
(a) Al₂(SO₄)₃

33. If equimolar solutions of urea, NaCl, sucrose and BaCl₂ have boiling points A, B, C and D, then
(a) A = C < B < D
(b) A = D < B < C
(c) A > B > C < D
(d) A < B < C < D
Answer:
(a) A = C < B < D

34. The molar mass of acetic acid obtained by measuring depression in freezing point is 115.8 gmol^-1. Hence the degree of association is
(a) 0.482
(b) 0.964
(c) 0.883
(d) 1.12
Answer:
(b) 0.964

35. Δ T_b/m for NaBr solution will have value (K_b = 0.52 K kg mol^-1)
(a) 0.52 K mol^-1
(b) 0.104 K kg mol^-1
(c) 1.24 kg mol^-1
(d) 1.04 K kg mol^-1
Answer:
(d) 1.04 K kg mol^-1

36. 0.2 M urea solution can be isotonic with
(a) 0.1 M KBr solution
(b) 0.1 M glucose solution
(c) 0.15 m NaCl solution
(d) 0.2 M KBr solution
Answer:
(a) 0.1 M KBr solution

37. The osmotic pressure of 0.1 M HCl solution at 27 °C will be
(a) 2.46 atm
(b) 0.164 atm
(c) 4.92 atm
(d) 0.0082 atm
Answer:
(c) 4.92 atm

38. If a, b, c and d are the van’t Hoff factors for Na₂SO₄, glucose and K₄[Fe(CN)₆] then
(a) a > b > c
(b) a < b < c
(c) b < a < c
(d) c < a < b
Answer:
(c) b < a < c

39. The osmotic pressure of 0.2 M KCl solution at 310 K is
(a) 10.17 atm
(b) 5.084 atm
(c) 8.36 atm
(d) 12.2 atm
Answer:
(a) 10.17 atm

40. A temperature at which 0.1 M KCl solution will have osmotic pressure 10 atm will be
(a) 408 °C
(b) 263 °C
(c) 310 °C
(d) 337 °C
Answer:
(d) 337 °C

41. 5 % solution of glucose is isotonic with a solution of urea (M = 60). Hence the weight of urea present in the solution is
(a) 1.67 g
(b) 6.0 g
(c) 18.6 g
(d) 1.2 g
Answer:
(a) 1.67 g

42. Abnormal molar mass is produced by
(a) association of solute
(b) dissociation of solute
(c) both association and dissociation of solute
(d) separation by semipermeable membrane
Answer:
(c) both association and dissociation of solute

43. The van’t Hoff factor for an aqueous solution of an electrolyte is
(a) less than 1
(b) zero
(c) greater than 1
(d) equal to 1
Answer:
(c) greater than 1

44. The value of van’t Hoff factor will be minimum for
(a) 0.05 M AlCl₃
(b) 0.2 M NaNO₃
(c) 5.0 M glucose
(d) 0.1 M H₂SO₄
Answer:
(c) 5.0 M glucose

45. van’t Haff factor for K₄[FeC(N)₆] dissociated 10% is
(a) 1.1
(b) 1.4
(c) 0.86
(d) 1.6
Answer:
(b) 1.4

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 1 Solid State Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 1 Solid State

Question 1.
What are the physical states of matter? How can they be changed into one another?
Answer:

• There are three physical states of matter namely solid, liquid and gas.
• They differ in intermolecular or interatomic or interionic forces which are strongest in the solid-state.
• By raising the temperature of solids to their melting point, solids are converted into liquids while heating liquids to their boiling points, they can be converted into vapour or gaseous state.
• On the contrary, by cooling the gases to very low temperature and subjecting to high pressure they can be transformed into liquid which on further cooling can be transformed into solid-state.

The equilibrium existing between three states of matter may be represented as,

Question 2.
What are the constituents of solids?
Answer:
The smallest constituent particles of various solids are atoms, ions or molecules. All such small constituents are referred to as ‘particles’.

Question 3.
What are the characteristic properties of solids?
Answer:

• The solid state of matter is characterised by strong interparticle forces of attraction.
• There is regularity and periodicity in the arrangement of constituent particles of solid.
• Generally solids are hard, incompressible and rigid except some solids like Na, K, P which are soft.
• The constituent particles of solids like molecules, atoms or ions have fixed stationary positions in solid and can only oscillate about their equilibrium or mean positions. Hence, they have fixed shape and cannot be poured like liquids.
• Crystalline solids have sharp melting points and they melt at a definite temperature. Amorphous solids do not have sharp melting points.
• They are anisotropic or isotropic.

Question 4.
Give classification of solids.
Answer:
Depending on orderly arrangement of the constituent particles, the solids are classified into two types :

• Crystalline solids. For example, diamond, NaCl, K₂SO₄, etc.
• Amorphous solids or non-crystalline solids. For example, tar, glass, plastics, rubber, butter, etc.

Question 5.
Define :
(1) Crystalline solid.
(2) Amorphous solid.
Answer:
(1) Crystalline solid : A homogeneous solid in which the constituent particles like atoms, ions or molecules are arranged in a definite repeating pattern throughout the solid is called crystalline solid. For example, NaCl, KNO₃, etc.

(2) Amorphous solid : A substance which appears like solid but does not have perfectly ordered crystalline structure and no regular arrangement of constituent particles in structure is called amorphous solid. For example, glass, rubber, plastics, etc.

Question 6.
Define the term anisotropy.
OR
Define and explain the term anisotropy.
Answer:
Anisotropy : The ability of crystalline solids to change their physical properties when measured in different directions is called anisotropy.

Explanation : This property is due to different arrangement of constituents in different directions. Different types of particles fall on the way of measurements in different directions. Hence the composition of crystalline solid changes with directions changing their physical properties.

Fig. 1.1 : Anisotropy in crystals : Different arrangements of constituent particles about different directions, AB, CD and EF.

Question 7.
Define and explain isotropy.
Answer:
Isotropy : The ability of amorphous solids to exhibit identical physical properties even though measured in different directions is called isotropy.

Explanation : This property arises because there is no long range order of regular pattern of arrangement in them and hence the arrangement is irregular along all the directions. Therefore the magnitude of any physical property would be identical along all directions.

Question 8.
Why does crystalline solid show different refractive indices in different directions ?
Answer:

1. Crystalline solid has long range order of regular pattern of arrangement which repeats periodically over entire crystal.
2. Within the given pattern, the arrangements of different atoms or ions or molecules is different in different directions. Hence the properties like refractive indices in the different directions are different.

This shows that the crystalline solids are anisotropic in nature.

Question 9.
Explain the properties of amorphous solids.
Answer:

1. The constituent particles in amorphous solids are arranged randomly.
2. They have short range ordered structure.
3. Amorphous solids are called supercooled liquids having very high viscosity.
4. They do not have sharp melting points and they melt gradually over a temperature interval.
5. Amorphous substances appear like solids but they do not have perfectly ordered crystalline structure, hence they are not real solids. Therefore they are pseudo solids.
6. They are isotropic and exhibit the same magnitude of any property in every direction.

Question 10.
What is isomorphism ?
Answer:
Isomorphism : A phenomenon in which two or more crystalline substances show same crystalline structure is called isomorphism and the crystals are said to be isomorphous. For example, NaNO₃ and CaCO₃. They have atomic ratios 1 : 1 : 3.

Question 11.
What is isomorphous?
Answer:
Isomorphous : When two or more crystalline substances have the same crystalline structure, they are said to be isomorphous. For example, NaF and MgO, NaNO₃ and CaCO₃.

Question 12.
What is polymorphism ?
Answer:
Polymorphism : A phenomenon in which when a single substance crystallises in two or more forms under different conditions of solidification is called polymorphism and the substance is called polymorphous. For example, calcite and oragonite are two forms of CaCO₃.

Question 13.
What is polymorphous ?
Answer:
Polymorphous: A single substance which crystallises in two or more forms under different conditions of solidification is called polymorphous. Polymorphic forms of an element are called allotropic forms or allotropes. For example, carbon exists as diamond and graphite, or sulphur exists in rhombic and monoclinic allotropic forms.

Question 14.
Identify isomorphous and polymorphous substances in the following :
K₂SO₄, graphite, β-quartz, Na₂SeO₄, CaCO₃, diamond, cristobalite, C_sNO₃.
Answer:

Question 15.
Why does a crystalline solid has a sharp melting point ?
Answer:

1. Crystalline solid is a homogeneous solid and it has long range order of regular pattern of arrangement which repeats periodically over entire crystal.
2. The interatomic or intermolecular forces are identical, hence the thermal energy required to break the regular structure by overcoming the intermolecular forces is uniform throughout.
3. Hence the heat and temperature needed to melt the solid are same, and therefore solids have sharp melting points.

Question 16.
Amorphous solids do not have sharp melting points. Explain.
Answer:

1. Amorphous solids do not have perfectly ordered crystalline structure.
2. They have short range order of regular pattern hence periodically repeating regular pattern is over a short distance.
3. The thermal energy required to break the structure and separate constituent particles is not uniform.
4. Hence the temperature needed to melt the solid is not same, therefore amorphous solids do not have sharp melting points but melt over a range of temperature.

Question 17.
Give examples of (1) crystalline solids and (2) amorphous solids.
Answer:
(1) Crystalline solids : Metallic solids (Cu, Fe, etc.) crystalline salts (NaCl, K₂SO₄, etc.)
(2) Amorphous solids : Glass, plastics, rubber, etc.

Question 18.
Distinguish between crystalline solids and amorphous solids.
Answer:
Crystalline solids:

1. They have definite characteristic geometrical shape.
2. They have long range order of regular pattern of arrangement of constituent particles.
3. They are true solids.
4. They have sharp melting points.
5. They are anisotropic in nature.
6. They have definite heat of fusion.

Amorphous solids:

1. They have irregular shape.
2. They have short range order of regular pattern of arrangement of constituent particles.
3. They are pseudo solids or supercooled liquids.
4. They do not have sharp melting points.
5. They are isotropic in nature.
6. They do not have definite heat of fusion.

Question 19.
How are crystalline solids classified ?
Answer:
Crystalline solids are classified as follows :

1. Ionic crystals
2. Covalent network crystals
3. Molecular crystals
4. Metallic crystals

Question 20.
What are crystalline solids?
Answer:
The solids in which the constituent particles are charged ions namely cations and anions held together by electrostatic force of attraction are called crystalline solids.

Question 21.
What are the characteristics of ionic crystals ?
Answer:
The characteristics of ionic crystals are as follows :

1. The constituents of ionic crystals are charged ions namely cations and anions. They differ in ionic size.
2. The ions in these crystals are held by strong electrostatic force of attraction.
3. Ionic crystals have high melting points and they are hard and brittle.
4. In solid state they are nonconductors of electricity but they are good conductors when melted or dissolved in water.
5. In aqueous solution they dissociate forming ions.
6. Example : NaCl, KCl. CaF₂, K₂SO₄, etc.

Question 22.
Explain why ionic solids are hard and brittle.
Answer:

1. In ionic crystalline solids, constituent particles are positively charged cations and negatively charged anions placed at alternate lattice points.
2. The ions are held by strong coulombic electrostatic forces of attraction compensating opposite forces. Hence they are hard.
3. Since there are no free electrons, they are not malleable and on applying a shearing force, ionic crystals break into small units. Hence they are brittle.

Question 23.
What are covalent network crystals ?
Answer:
The crystals in which the constituent particles are atoms linked by covalent bonds forming a continuous network are called covalent network crystals. For example, diamond, quartz.

Question 24.
What are the characteristics of covalent network crystals ?
Answer:
The characteristics of covalent network crystals are as follows :

• The constituent particles in these solids are atoms.
• The atoms in these crystals are held by covalent bonds forming a rigid three dimensional network which gives a giant molecule. Hence, the entire crystal is a single molecule.
• These crystals are very hard (or hardest) and most incompressible.
• They have high melting points and boiling points.
• Since the electrons are localised they are poor conductors of heat and electricity.
• Example : Quartz (SiO₂), diamond.

Question 25.
Give the examples of network solids.
Answer:
The examples of covalent network solids are as follows :
Quartz (SiO₂), diamond, boron nitride carborundum.

Question 26.
What are allotropes ?
Answer:
Allotropes : When a substance exists in two or more forms then they are called allotropes. They are polymorphous. For example, carbon has allotropes diamond and graphite.

Question 27.
What are molecular crystals ?
Answer:
The crystals in which the constituent particles are molecules (or unbonded single atoms) of the same substance held together by intermolecular forces of attraction. For example solidified Cl₂, CO₂, etc.

Question 28.
What are the characteristics of molecular crystals ?
Answer:
The characteristics of molecular crystals are as follows :

1. The constituent particles of these solids are molecules (or unbonded single atoms) of the same substance.
2. The atoms within the molecules are bonded by covalent bonds.
3. The molecules are held together by intermolecular forces of attraction.

Question 29.
What are intermolecular forces of attraction involved in molecular crystals ?
Answer:
The intermolecular forces involved in molecular crystals are as follows :
(1) Weak dipole-dipole interactions :
The solids constituting polar molecules like HCl, H₂O, SO₂, etc. which possess permanent dipole moment involve weak dipole-dipole interactions.

(2) Very weak dispersion or London forces :
The solids consisting of nonpolar molecules like CH₄, H₂, etc. involve weak dispersion forces. They are also involved in monoatomic solids like Ar, Ne.

(3) Intermolecular Hydrogen bonds :

• In this crystalline solids, the constituent particles are the molecules which contain hydrogen atom linked to highly electronegative atom like F, O or N.
• In these, molecules are held by hydrogen bonds in which H atom of one molecule is bonded to electronegative atom (like F, N or O) of another molecule.
• Since hydrogen bonding is weak, these solids have very low melting points and generally at room temperature they exist in the liquid or gaseous state.
• They are non-conductors of electricity.

Question 30.
What are metallic crystals ?
Answer:
These are crystalline solids formed by atoms of the same metallic element held together by metallic bonds.

Question 31.
What are the characteristics of metallic crystals ?
Answer:

1. Metallic crystals are solids formed by atoms of the same metallic element held together by metallic bonds.
2. Metallic crystals have high melting point and boiling point.
3. Metals are malleable and can be hammered into thin sheets.
4. Metals are ductile and can be drawn into thin wires.
5. Metals are good conductors of heat and electricity.
6. Examples are Cu, Ag, Au, Ni, etc.

Question 32.
Classify the following solids into different types :
(i) Plastic (ii) P₄ molecule (iii) S₈ molecule (iv) Iodine molecule (v) Tetra phosphorus decoxide (vi) Ammonium phosphate (vii) Brass (viii) Rubidium (ix) Graphite (x) Diamond (xi) NaCl (xii) Silicon.
Answer:

Question 33.
Mention the types of the following solids :
(i) CaF₂ (ii) SiC (iii) Ice (iv) SO₂ (v) CaCO₃ (vi) ZnS (vii) HCl (viii) CO₂
Answer:

Question 34.
What is a giant solid ?
Answer:
Covalent solid formed by covalent bonds between neighbouring constituent atoms of non-metallic solid is called a giant solid. For example, graphite.

Question 35.
What is lattice ?
Answer:
Lattice is a geometrical arrangement of points in a three dimensional periodic array.

Question 36.
What is crystal lattice (space lattice) ?
Answer:
Crystal lattice (space lattice) : A regular arrangement of the constituent particles (atoms, ions or molecules) of a crystalline solid having similar environment in three dimensional space is called crystal lattice or space lattice.

Question 37.
What is a lattice point?
Answer:
Lattice point : A position occupied by a crystal constituent particle like an atom, ion or a molecule in the crystal lattice is called lattice point or lattice site.
OR
Any point at the intersection of the lines in the unit cell occupied by a constituent particle like an atom, an ion or a molecule in the crystalline solid is called a lattice point.

Question 38.
What are the parameters of a unit cell ?
Answer:
A unit cell is characterised by following parameters :
(1) Edges or edge lengths : The intersection of two faces of crystal lattice is called as edge. The three edges denoted by a, b and c represent the dimensions (lengths) of the unit cell along three axes. These edges may or may not be mutually perpendicular.

(2) Angles between the edges (or planes) : There are three angles between the edges of the unit cell represented as α, β and γ.

• The angle α is between edges b and c.
• The angle β is between edges a and c.
• The angle γ is between edges a and b.

The crystal is defined with the help of these parameters of its unit cell.

Question 39.
Represent space lattice and unit cell diagrammatically.
Answer:

Question 40.
What do you understand by the basis of crystal lattice ?
Answer:

• A crystal structure is formed by attaching a constituent particle to lattice points.
• The constituent particles attached to the lattice points form the basis of the crystal lattice.
• The crystal structure is obtained by attaching a basis to each of the lattice points.

This is represented by the following schematic equation :

Question 41.
What are the types of unit cells?
Answer:
Basically unit cells are of two types as follows :

1. Primitive unit cells : The unit cells in which the constituent particles like atoms, ions or molecules are present only at the corners of the unit cell are called primitive unit cells or simple unit cells.
2. Body-centred unit cell : A unit cell in which the constituent particles are present at the corners as well as at its body-centre is called body-centred unit cell.
3. Face-centred unit cell : A unit cell in which the constituent particles are present at the corners as well as at the centre of each face is called face-centred unit cell or cubic close packed (CCP) unit cell.
4. Base-centred unit cell : A unit cell in which the constituent particles are present at the corners as well as at the centres of two opposite faces is called end-centred unit cell.

Question 42.
Explain briefly crystal systems.
Answer:
(1) The constituent particles like atoms, ions or molecules of the crystal can be arranged in seven different ways changing edges (a, b, c) and angles (α, β, γ) and accordingly they form seven systems or types of the crystal.

(2) These seven crystal system are named as :
(a) Cubic system, (b) Tetragonal system (c) Orthorhombic system (d) Rhombohedral system (e) Monoclinic system (f) Triclinic system (g) Hexagonal system.

Question 43.
What are Bravais lattices ?
Answer:

1. There are seven crystal systems according to the edges (a. b, c) and angles (α, β, γ).
2. The constituents of the crystal may be present at corners, face centres, body centres, edge centres and voids.
3. By mathematical analysis, it has been proved that only fourteen different kinds of space lattices are possible.
4. Hence there are fourteen different ways of arrangement of the lattice basis.
5. These fourteen lattices of seven crystal systems are called Bravais lattices.

Question 44.
Explain Bravais lattices of a cubic system.
OR
Explain unit cells of a cubic system.
Answer:
Cubic lattice : For this, edges are a = b = c and angles are α = β = γ = 90°. In this cubic system, there are three Bravais lattices.
(1) Simple (or primitive) cubic unit cell (SCC) : In this unit cell, atoms are present only at 8 corners of the cube.

(2) Body-centred cubic unit cell (BCC) : In this, atoms are present at 8 corners along with one additional atom at the body-centre of the cube.

(3) Face-centred cubic unit cell (FCC) : In this unit cell, atoms are present at 8 comers and at 6 face centres.

Question 45.
Give the number of lattice points in one unit cell of the following crystal structures :
(1) Simple cubic
(2) Face-centred cubic
(3) Body-centred cubic.
Answer:
Lattice points in one unit cell represent the positions of atoms, ions or molecules in the unit cell.
(1) Simple cubic unit cell : In this primitive unit cell, the lattice points are at 8 corners of the unit cell. Hence there are 8 lattice points.
(2) Face-centred cubic unit cell : In this unit cell, the lattice points are at 8 comers and 6 face centres.
(In cubic close packing unit cell, the lattice points are also at edge centres and body centre.)
(3) Body-centred cubic unit cell : In this, the lattice points are at 8 comers and one at body centre.

Question 46.
Find the number of atoms per unit cell in the following crystal structures.
(1) Simple cubic unit cell
(2) Body-centred cubic unit cell
(3) Face-centred cubic cell.
Answer:
(1) Number of atoms in primitive simple cubic (scc) unit cell :

In simple or primitive cubic unit cell, there are 8 atoms at 8 corners. Each corner contributes 1/8th atom to the unit cell.
∴ Number of atoms present in the unit cell = \(\frac {1}{8}\) × 8 = 1
Hence the volume of the unit cell is equal to the volume of one atom.

(2) Number of atoms in body-centred cubic (bcc) unit cell:

In this unit cell, there are 8 atoms at 8 corners and one additional atom at the body centre. Each corner contributes 1/8th atom, to the unit cell, hence due to 8 corners.
Number of atoms = 8 × \(\frac {1}{8}\)
= 1 atom.
An atom at the body centre wholly belongs to the unit cell.
∴ Total number of atoms present in bcc unit cell = 1 + 1 = 2.
Hence the volume of unit cell is equal to the volume of two atoms.

(3) Number of atoms in face-centred cubic (fcc) unit cell :

In this unit cell, there are 8 atoms at 8 comers and 6 atoms at 6 face centres. Each corner contributes 1/8th atom to the unit cell, hence due to 8 corners,
Number of atoms = \(\frac {1}{8}\) × 8 = 1.
Each face centre contributes half of the atom to the unit cell, hence due to 6 face centres,
Number of atoms = \(\frac {1}{2}\) × 6 = 3.
∴ Total number of atoms present in fee unit cell = 1 + 3 = 4.
Hence the volume of the unit cell is equal to the volume of four atoms.

Question 47.
Obtain a relation for the density of the unit cell and radius of atom or sphere for the following :
(1) Simple cubic (scc) crystal
(2) Body centred cubic (bcc) crystal
(3) Face centred cubic (fcc) crystal.
Answer:
(1) Consider a unit cell of a simple cubic crystal. It has 8 atoms at 8 corners of the unit cell.
∴ Total number of atoms in unit cell = \(\frac {1}{8}\) × 8 = 1
If a is the length of edge of cubic unit cell and r is the radius of the atom, then r = a/2 or a = 2r.
Volume of the unit cell = a³ = (2r)³ = 8r³
If M is atomic mass of the element, then mass of one atom is M/N_A where N_A is Avogadro number. If there are V atoms in one unit cell then,
Mass of unit cell = n × Mass of one atom = n × \(\frac{M}{N_{\mathrm{A}}}\)

Since there is one atom present in one unit cell,
ρ = \(\frac{M}{N_{\mathrm{A}} \times a^{3}}\)

(2) Consider a unit cell of body centred cubic (bcc) crystal. It has 8 atoms at 8 comers and one additional atom at the centre of body of unit cell.
Number of atoms due to 8 corners = \(\frac {1}{8}\) × 8 = 1
Body centred atom, wholly belong to the unit cell. Hence total number of atoms in the unit cell is two. If M is atomic mass of an element then M/N_A is mass of one atom where N_A is Avogadro number.
Mass of unit cell = Mass of 2 atoms in unit cell = 2 M/N_A
If a is the edge length or lattice parameter then,
Volume of cubic unit cell = a³

(3) Consider a unit cell of face centred cubic (fcc) crystal.
It has 8 atoms at 8 comers and 6 atoms at 6 face centres.
∴ Total number of atoms in unit cell = \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6 = 1 + 3= 4
If M is the atomic mass of an element, then mass of one atom is M/N_A, where N_A is Avogadro number. Mass of unit cell = Mass of 4 atoms
= \(4 \times \frac{M}{N_{\mathrm{A}}}\)
If a is the edge length of this cubic unit cell then,
volume of unit cell = a³

Question 48.
What is coordination number? What is its significance?
Answer:
(1) Coordination number : The number of the closest neighbouring constituent particles like atoms, ions or molecules which are in contact with a particular particle or an atom in the crystal lattice is called coordination number of that particle.
(In the crystal lattice, all atoms may have same or different coordination numbers.)

(2) The magnitude of the coordination number is a measure of compactness of spheres in close-packed structures.

(3) The higher the coordination number, the closer are the spheres to each other.

Question 49.
Explain linear packing in one direction.
OR
Explain close packing in one dimension.
Answer:
Linear packing in one direction or close packing in one dimension :
The constituent particles of the crystal may be of varying shapes but for better understanding we consider particles as hard spheres of equal size.

There is only one way of arranging or packing spheres placed in a horizontal row touching one another. Since one sphere is in contact with two neighbouring spheres except the end atoms, the coordination number in this arrangement is two. This packing may be in any one direction x, y or z.

Question 50.
Explain the following :
Planar packing arrangement of spheres.
OR
Close packing in two dimensions.
OR
AAAA type and ABAB type of two dimensional arrangement.
OR
(i) Square close packing
(ii) Hexagonal close packing.
Answer:
Two dimensional close packing crystal structure can be generated by placing one dimensional linear crystal structure over another to form multiple layers. This staking of linear rows may be taking place in two different ways giving two different two dimensional structures as follows :

(i) AAAA type two dimensional close packing or square close packing :
In this arrangement, various one dimensional rows are placed on one over other so that each sphere in one row is over the another sphere of another row forming planar structure. In this, spheres have horizontal as well as vertical alignment. All the rows of spheres are identical in planar structure. All crests as well as all the depressions or troughs formed by the arrangement are also aligned.

If the first row is labelled as A type, then second and all subsequent rows are also identical, hence are of A type. Therefore this planar two dimensional close packing is called AAAA type packing.

In this arrangement, each sphere is in contact with (or touching) four other spheres around it, hence the coordination number of each sphere is four and the packing is called two dimensional or planar square close packing. In this, packing efficiency is 52.4%.

(ii) ABAB type two dimensional packing or hexag-onal close packing :

In this arrangement, crests of the spheres of one row are placed into the depressions or troughs formed between adjacent spheres of next row. This arrangement is repeated consecutively throughout.

In this arrangement, crests of the spheres of one row are in contact with depressions or troughs of next row.

If one row of spheres is labelled as A then the next row will be B, third row will again be A, fourth row B and so on. Hence this planar or two dimensional close packing is called ABAB… type packing.

In this arrangement, each sphere is in contact with six other spheres around it hence the coordination number of each sphere is six and the packing is called two dimensional or planar hexagonal close packing. In this, the packing efficiency is 60.4% which is more than linear close packing.

Question 51.
Explain close packing in three dimensions.
Answer:
Close packing in three dimensions :
Three dimensional crystal structures are obtained by stacking of two dimensional layers. Simple cubic lattice is obtained by stacking of two dimensional square layers.

The stacking of two dimensional hexagonal close packed layers gives two structures namely hexagonal close packed (hcp) structure and face centred (fcc) structure.

(i) Stacking of square close packed layers :

In this arrangement, the two dimensional AAAA type square closed packed layers are placed one over the other in such a way that the crests of all spheres are in contact with successive layers in all directions. All spheres of different layers are perfectly aligned horizontally and vertically forming unit cells having primitive or simple cubic structure. Since all the layers are identical and if each layer is labelled as layer A, then whole three dimensional crystal lattice will be of AAAA… type.

Each sphere is in contact with six surrounded spheres, hence the co-ordination number of each sphere is six.

(ii) Stacking of two hexagonal close packed layers :
A close packed three dimensional structure can be generated by arranging hexagonal close packed layers in a particular manner.

In this the spheres of second layer are placed in the depression of the first layer.

In this if first layer is labelled as A then second layer is labelled as B since they are aligned differently.

In this, all triangular voids of the first layers are not covered by the spheres of the second layer. The triangular voids which are covered by second layer spheres generate tetrahedral void which is surrounded by four spheres. The triangular voids in one layer have above them triangular voids of successive layers.

The overlapping triangular voids from two layers together form an octahedral void which is surrounded by six spheres.

(iii) Placing third hexagonal close packed layer :
(a) Hexagonal close packing (hcp) : If the crests of spheres of third layer are placed on the triangular shaped tetrahedral voids C of the second layer, then three dimensional closest packing structure is obtained in which the spheres of third layer lie directly above the spheres of first layer, i.e., first and third layers are identical. Following same placing of layers, fourth layer will be identical to second layer.

If the first layer is labelled A and second layer B, then the arrangement of packing will be of ABAB type. This is also called hexagonal close packing (hcp) as shown in the figure. In this, packing efficiency is 74%. The coordination number of each sphere is 12. The metals Be. Mg, Zn, Cd crystallise in HCP crystalline structure.

(b) Cubic close packing (ccp) : If the crests of spheres of third layer are placed in the positions of tetrahedral void ‘a’ having apex upward of first layers, then the third layer will not be identical to the first, and may be labelled as C, which is different from A and B layers. Fourth layer may be arranged above third layer such that the spheres are aligned, so that the first layer and fourth layer are identical, second and fifth layers are identical and so on.

If first, second and third layers are labelled as A, B and C respectively then the arrangement of packing will be ABCABC type. This is also called cubic close packing (ccp) as shown in the figure. This is similar to face centred cubic (fcc) packing.

In this, arrangement packing efficiency is 74% and the coordination number of each sphere is 12.

Question 52.
Explain tetrahedral void.
Answer:
(1) Tetrahedral void : The vacant space or void among four constituent particles having tetrahedral arrangement in the crystal lattice is called tetrahedral void.

The arrangement of four spheres around the void is tetrahedral. A tetrahedral void is formed when a triangular void made by three coplanar spheres is in contact with fourth sphere above or below it.

(2) Characteristics of tetrahedral void :

• The volume of the void is much smaller than that of atom or sphere.
• Larger the size of sphere, more is the size of void.
• If R is the radius of the constituent atom, then the radius of the tetrahedral void is 0.225 R.
• Coordination number of tetrahedral void is four.
• There are two tetrahedral voids per sphere, in the crystal lattice. If the number of closed packed spheres is N then the number of tetrahedral voids is 2N.

Question 53.
Explain octahedral void.
Answer:
(1) Octahedral void : The vacant space or void at the centre of six spheres (or atoms) which are placed octahedrally is called octahedral void.

(2) Characteristics of octahedral void :

• The volume of the void is small.
• There is one octahedral void at the body centre and twelve octahedral void positions at twelve edge centres.
• If R is the radius of constituent atom, then the radius of the octahedral void is 0.414 R.
• The coordination number of octahedral void is six.
• There is one octahedral void per sphere in the crystal lattice. If the number of closed packed spheres is N then the number of octahedral voids is N.

Question 54.
What are number of voids per atom in hep and ccp ?
Answer:
The tetrahedral and octahedral voids occur in hep and ccp/fcc structures. There are two tetrahedral voids associated with each atom. The number of octahedral voids is half that of tetrahedral voids. Thus, there is one octahedral void per atom.

Question 55.
What is packing efficiency?
Answer:
(1) Packing efficiency : It is the fraction of a percentage of the total space (of the unit cell) occupied by the particles (spheres).

( 2) The magnitude of packing efficiency gives a measure of how tightly particles are packed together.

Question 56.
Calculate packing efficiency in body-centred cubic lattice.
Answer:
Step 1 : Radius of sphere :
In the unit cell of body-centred cubic lattice, there are 8 atoms at 8 corners and one atom at the centre of the cube.

The atoms are in contact along the body diagonal BF. Let a be the edge length and r the radius of an atom.
Consider a triangle BCE.
BE² = BC² + CE² = a² + a² = 2a²
Consider triangle BEF,
BF² = BE² + EF² = 2a² + a² = 3a²
BF = \(\sqrt{3}\)a.
From figure, BF = 4r
∴ 4r = \(\sqrt{3}\)a
∴ r = \(\frac{\sqrt{3}}{4} a\)

Step 2 : Volume of sphere :

Step 3 : Total volume of particles :
The unit cell of bcc structure contains 2 particles.

Step 4 : Packing efficiency

∴ Packing efficiency = 68%
∴ Percentage of void space = 100 – 68
= 32%

Question 57.
Calculate packing efficiency in face-centred cubic lattice.
Answer:
Step 1 : Radius of sphere :
In the unit cell of face-centred cubic lattice, there 8 atoms at 8 corners and 6 atoms at 6 face centres. Consider the face ABCD.

The atoms are in contact along the face diagonal BD.
Let a be the edge length and r, the radius of an atom.
Consider a triangle BCD.
BD² = BC² + CD²
= a² + a² = 2a
∴ BD = \(\sqrt{2} a\)
From figure, BD = 4r

Step 3 : Total volume of particles :
The unit cell of fee crystal lattice contains 4 particles.
∴ Volume occupied by 4 particles = \(4 \times \frac{\pi a^{3}}{12 \sqrt{2}}\)
= \(\frac{\pi a^{3}}{3 \sqrt{2}}\)

Step 4 : Packing efficiency :

∴ Packing efficiency = 74%
∴ Percentage of void space = 100 – 74
= 26%

Edge length and particle parameters in cubic system

Coordination number and packing efficiency in systems

Question 58.
Calculate the number of particles and unit cells in ‘x’ g of metallic crystal.
Answer:
Consider ‘x’ gram of a metallic crystal of molar mass M and density ρ. If the unit cell of the crystal has edge length ‘a’ then, volume of unit cell = a³.
Mass of one metal atom = \(\frac{M}{N_{\mathrm{A}}}\). If ‘n’ number of atoms are present in one unit cell then
Mass of unit cell = \(\frac{n \times M}{N_{\mathrm{A}}}\)
If ‘a’ is the edge length of the unit cell then, Volume of unit cell = a³
Density of unit cell = ρ = \(\frac{n \times M}{N_{\mathrm{A}}} \times \frac{1}{a^{3}}\)
∴ M = \(\frac{\rho \times N_{\mathrm{A}} \times a^{3}}{n}\)
∵ Molar mass M gram contains N_A particles
∴ x gram contains \(\frac{x \times N_{\mathrm{A}}}{M}\) particles.
Substituting the value M,

(i) Number of particles in x g crystal
= \(\frac{x \times N_{\mathrm{A}}}{\rho \times N_{\mathrm{A}} \times a^{3} / n}\)
= \(\frac{x \times n}{\rho \times a^{3}}\) particles

(ii) Number of unit cells in x g crystal :
∵ n particles are present in 1 unit cell
∴ \(\frac{x \times n}{\rho \times a^{3}}\) are present in, \(\frac{x \times n}{\rho \times a^{3}} \times \frac{1}{n}\)
= \(\frac{x}{\rho \times a^{3}}\) unit cells

(iii) Number of unit cells in V volume of crystal = \(\frac{V}{a^{3}}\)

Alternative method :
Consider ‘x’ g metal of atomic mass M g mol^-1.
Number of moles of metal = \(\frac{x}{M}\)
(a) Number of atoms (particles) of metal = \(\frac{x}{M} \times N_{\mathrm{A}}\)
(b) If unit cell contains ‘n’ atoms,
Number of unit cells = \(\frac{x}{M} \times \frac{N_{\mathrm{A}}}{n}\)
(c) If ‘a’ is the edge length then,
Volume of unit cells = a³
∴ Number of unit cells in V volume of crystal = \(\frac{V}{a^{3}}\).

Solved Examples 1.5 – 4.7

Question 59.
Solve the following :
[1 m = 10 dm = 100 cm = 10⁹ nm = 10¹² pm, 1Å = 10^-8 cm = 100 pm]

(1) A cubic unit cell of a crystal consists of atoms of A and B elements. Atoms of A occupy corners of the unit cell while one B atom is present at the body centre. Determine the formula of the crystalline compound.
Solution :
Given : Atoms of A are at the comers while atom B is at the body centre of the cubic unit cell.
Since \(\frac {1}{8}\)th atom is contributed at each comer and there are 8 comers in unit cell, number of atoms of A due to comers is \(\frac {1}{8}\) × 8 = 1.

In addition there is one atom of B at body centre. Hence this unit cell contains one atom of each, A and B, therefore the formula of the compound is AB.
Ans. The formula of the compound = AB.

(2) Atoms C and D form fee crystalline structure. Atom C is present at the corners of the cube and D is at the face centres of the cube. What is the formula of the compound ?
Solution :
Given : Crystal has fee structure.
Atoms C are at 8 comers while atoms D are at 6 face centres of the cubic unit cell.
At the comer, \(\frac {1}{8}\)th of each C atom is present while at each face centre, half of each D atom is present.
Number of C atoms = \(\frac {1}{8}\) × 8 = 1.
Number of D atoms = \(\frac {1}{8}\) × 6 = 3.
Thus unit cell contains one C atom and three D atoms.
Hence the formula of the compound is CD₃.
Ans. Formula of the compound = CD₃.

(3) A cubic unit cell contains atoms A at the corners, atoms B at face centres and atom C at the body centre. What is the formula of the crystalline compound?
Solution :
Given : Atoms A are at 8 comers, atoms B at the 6 face centres and one atom C at body centre.
Total number of atoms of A = \(\frac {1}{8}\) × 8 = 1.
Total number of atoms of B = \(\frac {1}{2}\) × 6 = 3.
One atom of C at the body centre.
Therefore the unit cell contains one atom of A, three atoms of B and one atom of C.
Hence the formula of the compound is AB₃C.
Ans. The formula of the crystalline compound is AB₃C.

(4) An element A and B constitute bcc type crystalline structure. Element A occupies body centre position and B is at the corners of cube. What is the formula of the compound? What are the coordination numbers of A and B ?
Solution :
Given : Crystalline structure is bcc type.
Atoms A are at 8 comers and atom B is at body centre.
∴ Number of atoms of A in a unit cell
= \(\frac {1}{8}\) × 8 = 1.
Number of atom B in a unit cell = 1.
Since unit cell contains one atom each of A and B, the formula of the compound is AB.
The coordination number of an atom A at comer is 8.
The coordination number of an atom B at body centre is 8.
Answer. Formula of the compound = AB.
Coordination number of A = 8
Coordination number of B = 8.

(5) Mention the number of atoms in the following unit cells :
(a) scc (b) bcc (c) fcc (d) hcp.
Answer:

(6) 0.1 mole of Buckminster fullerene of molar mass 720 gmol^-1 contains how many Kg of carbon ?
Solution :
Molar mass of fullerene, C60 = 720 gmol^-1
∵ Mass of 1 mole of fullerene = 720 g = 0.72 kg
∴ Mass of 0.1 mole of fullerene = 0.72 × 0.1
= 0.072 kg
Ans. 0.072 kg.

(7) In a cubic crystalline structure of zinc blende (ZnS), sulphide ions are at corners and face centres while zinc ions occupy half of tetrahedral voids.
Find in the unit cell :
(i) Number of Zn²⁺ ions
(ii) Number of S^2- ions
(iii) Number of ZnS molecules
(iv) Molecular formula of zinc blende.
Solution :
Given : S^2- ions are at 8 comers and 6 face centres.
Zn²⁺ ions occupy half of tetrahedral voids.
Number of S^2- ions in unit cell

= 1 + 3 = 4

Cubic unit cell has 8 tetrahedral voids. Since half of them are occupied by Zn²⁺ ions, there are 4Zn²⁺ ions in the unit cell.
Hence number of zinc sulphide (ZnS) molecules is 4.
Molecular formula of zinc blende is ZnS.
Ans. (i) Number of Zn²⁺ ions = 4
(ii) Number of S^2- ions = 4
(iii) Number of ZnS molecules = 4
(iv) Molecular formula of zinc blende = ZnS.

(8) In a crystalline compound, atoms A occupy ccp lattices while atoms B occupy 2/3 rd tetrahedral voids. What is the formula of the compound ?
Solution :
In ccp unit, lattice points are 8 comers and 6 face centres where atoms A are present.
∴ Number of A atoms

In cubic unit cell, there are 8 tetrahedral voids.
Hence,
Number of B atoms = \(\frac {2}{3}\) × 8 = \(\frac {16}{3}\).
Hence the formula should be A₄B_16/3 or A₁₂B₁₆ or A₃B₄.
Ans. The formula of the compound is A₃B₄.

(9) A compound forms hep structure. What is the number of (a) octahedral voids, (b) tetrahedral voids, (c) total voids formed in 0.2 mol of the compound?
Solution :
Number of atoms in 0.2 mol of the compound
= 0.2 × N_A = 0.2 × 6.022 × 10²³
= 1.2044 × 10²³ atoms
(a) Number of octahedral voids
= Number of atoms
= 1.2044 × 10²³

(b) Number of tetrahedral voids
= 2 × Number of atoms
= 2 × 1.2044 × 10²³
= 2.4088 × 10²³

(c) Total number of voids
= 1.2044 × 10²³ + 2.4088 × 10²³
= 3.6132 × 10²³
Ans. (a) Number of octahedral voids
= 1.2044 × 10²³
(b) Number of tetrahedral voids = 2.4088 × 10²³
(c) Total number of voids = 3.6132 × 10²³

(10) Copper crystallises into a fcc structure and the unit cell has length of edge 3.61 × 10^-8 cm. Calculate the density of copper. Atomic mass of copper is 63.5 g mol^-1.
Solution :
Given : Crystalline structure of Cu is fcc.
Edge length = a = 3.61 × 10^-8 cm
Atomic mass of Cu = 63.5 g mol^-1
Avogadro number = 6.022 × 10²³ mol^-1
Density = d = ?

In fcc structure, there are 8 Cu atoms at 8 comers and 6 Cu atoms at 6 face centres.
∴ Total number of Cu atoms
= \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6 = 1 + 3
= 4
Mass of one Cu atom
= \(\frac{63.5}{6.022 \times 10^{23}}\) = 1.054 × 10^-22 g
∴ Mass of 4 Cu atoms = 4 × 1.054 × 10^-22
= 4.216 × 10^-22 g
Mass of unit cell = Mass of 4 Cu atoms
= 4.216 × 10^-22g
Volume of unit cell = a3 = (3.61 × 10^-8)3
= 4.7 × 10^-24 cm³
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
∴ ρ = \(\frac{4.216 \times 10^{-22}}{4.7 \times 10^{-23}}\) = 8.97 g cm^-3
Ans. Density of Cu = 8.97 g cm^-3.

(11) Niobium is found to crystallise with bcc structure and found to have density of 8.55 g cm^-3 (OR 8.55 kg m^-3). Determine the atomic radius of niobium if its atomic mass is 93.
Solution :
Given : Density of Niobium (Nb) crystal = 8.55 g cm^-3
Crystalline stmeture is bcc.
Atomic mass of Nb = 93 g mol^-1
Avogadro number = N_A = 6.022 × 10²³ mol^-1
Atomic radius of Niobium = ?
In bcc unit cell, there are 8 atoms at 8 comers and 1 atom at the body centre.
∴ Number of Nb atoms = \(\frac {1}{8}\) × 8 + 1 = 1 + 1 = 2.
Mass of one Nb atom = \(\frac{93}{6.022 \times 10^{23}}\) = 1.544 × 10^-22
∴ Mass of 2 Nb atoms = 2 × 1.544 × 10^-22 = 3.088 × 10^-22 g
Mass of unit cell
= Mass of 2Nb atoms = 3.088 × 10^-22 g
If a is edge length of bcc unit cell, volume of unit cell = a³

= 0.361 × 10^-22 cm³ = 36.1 × 10^-24 cm³
∴ a = (36.1 × 10^-24)^-1/3 = 3.3 × 10^-8 cm
If r is the radius of 1 Nb atom, then in bcc structure
r = \(\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4}\) × 3.3 × 10^-8
= 1.43 × 10^-8 cm
= 1.43 × 10^-10 m
= 1.43 × 10^-10 × 10⁹ nm
= 0.143 nm
Ans. Atomic radius of niobium atom = 0.143 nm

(12) Gold occurs as face centred cube and has a density of 19.30 kg dm^-3. Calculate atomic radius of gold. (Molar mass of Au = 197)
Solution :
Given : Density of Au = 19.3 kg dm^-3
Molar mass = 197 g mol^-1
Avogadro constant = N_A = 6.022 × 10²³ mol^-1
Atomic radius of Au = ?
In fcc unit cell, there are 8 atoms of Au at 8 comers and 6 atoms at 6 face centres.
Number of Au atoms in the unit cell = \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6
= 4 atoms
Mass of 1 Au atom = \(\frac{197}{6.022 \times 10^{23}}\) = 3.271 × 10^-22 g
∴ Mass of 4 Au atoms = 4 × 3.271 × 10^-22 g = 1.308 g × 10^-21 g
∴ Mass of unit cell = 1.308 × 10^-21 g

= 6.777 × 10^-24 dm3
= 6.777 × 10^-23 cm3
∴ a = (6.777 × 10^-23)^1/3 = (67.77 × 10^-24)^-1/3
= 4.077 × 10^-8 cm
If r is the radius of Au atom, then for fcc unit cell,
r = \(\frac{a}{2 \sqrt{2}}\)
= \(\frac{4.077 \times 10^{-8}}{2 \sqrt{2}}\) = 1.442 x 10^-8 cm = 144.2 pm
Ans. Radius of Au atom = 144.2 pm.

(13) A compound is formed by two elements X and Y. The atoms of Y form ccp structure. The atoms of A occupy \(\frac {1}{3}\) of tetrahedral voids. Find the formula of the compound.
Solution :
In ccp structure, Y atoms are present at 8 comers and 6 face-centres of the ccp structure.

There are 8 tetrahedral voids of which \(\frac {1}{3}\)rd are occupied by atoms X. Hence unit cell has \(\frac {8}{3}\) atoms of X and 4 atoms of Y. The formula will be,
X_8/3 Y₄ or X₈Y₁₂ or X₂Y₃
Ans. Formula of the compound = X₂Y₃.

(14) A metal crystallises into two cubic faces namely face centered (FCC) and body centered (BCC), whose unit cell edge lengths are 3.5 Å and 3.0 Å respectively. Find the ratio of the densities of FCC and BCC.
Solution :
Given : Edge length of unit cell of fcc metal = 3.5 Å
= 3.5 × 10^-8 cm
Edge length of unit cell of bcc metal = 3 Å = 3 × 10^-8 cm
Density d = \(\frac{n \times \mathbf{M}}{a^{3} \times \mathbf{N}_{\mathrm{A}}}\)
where, n = Number of Fe atoms in the unit cell
M = Atomic mass of metal
a = Edge length of unit cell
N_A = Avogadro number
∴ For fcc unit cell = n = 4
For bcc unit cell = n = 2
∴ \(\frac{\text { Density of fcc unit cell }}{\text { Density of bcc unit cell }}=\frac{d_{\mathrm{fcc}}}{d_{\mathrm{bcc}}}\)

(15) The density of silver having atomic mass 107.8 gram mol^-1 is 10.8 gram cm^-3. If the edge length of cubic unit cell is 4.05 × 10^-8 cm, find the number of silver atoms in the unit cell. (N_A = 6.022 × 10²³, 1Å = 10^-8 cm)
Solution :
Given : d = 10.8 g cm^-3
M= 107.8 g mol^-1
a = 4.05 × 10^-8 cm
Number of Ag atoms in unit cell = n = ?
Mass of one Ag atom = \(\frac{107.8}{6.022 \times 10^{23}}\)
= 1.79 × 10^-22 g
If there are n atoms, then
Mass of unit cell = n × 1.79 × 10^-22 g
Volume of unit cell = a3 = (4.05 × 10^-8)³
= 6.643 × 10^-23 cm³

Ans. Number of silver atoms (Ag) atoms in unit cell = 4.

(16) Aluminium having atomic mass 27 g mol^-1 crystallises in face centred packed cubic crystal. Find the number of Al atoms in 10 g aluminium. How many unit cells will be present in it?
Solution :
Given : Atomic mass of Al = 27 g mol^-1
Mass of Al = 10 g
Avogadro number = N_A = 6.022 × 10²³ mol^-1
Number of Al atoms = ?
Number of unit cells = ?
1 gram atom of Al = 27 g Al contains 6.022 × 10²³
Al atoms
∴ Number of Al atoms in 10 g
= \(\frac{10 \times 6.022 \times 10^{23}}{27}\)
= 2.23 × 10²³
In fcc structure, each unit cell contains 4Al atoms.
∴ Number of unit cells = \(\frac{2.23 \times 10^{23}}{4}\)
= 5.575 × 10²²
Ans. Number of Al atoms = 2.23 × 10²³
Number of unit cells = 5.575 × 10²².

(17) The density of iron crystal is 8.54 gram cm^-3. If the edge length of unit cell is 2.8 Å and atomic mass is 56 gram mol^-1, find the number of atoms in the unit cell. What is the type of crystal ?
Solution :
Given :
Density of Fe crystal = d = 8.54 g cm^-3
a = 2.8 Å = 2.8 × 10^-8 cm
Atomic mass = M = 56 g mol^-1
Number of atoms in unit cell, n = ?
Mass of one atom = \(\frac{56}{6.022 \times 10^{23}}\) = 9.3 × 10^-23 g
Volume of unit cell = a³ = (2.8 × 10^-8)³
= 2.195 × 10^-23 cm³
If there are n atoms in the unit cell, then
Mass of unit cell = n × 9.3 × 10^-23 g

Since unit cell contains 2 atoms, the crystal has bcc structure.
Ans. Number of atoms in unit cell = 2
Type of crystal = bcc

Question 60.
What is meant by crystal defects or imperfections?
Answer:

1. Defect in crystalline structure : Any deviation from orderly and stoichiometrically perfect arrangement of atoms, ions or molecules in the crystal lattice is called a defect in the crystalline structure.
2. Defects are created during the crystallisation process. If the process occurs at faster rate, the defects are more.
3. The properties of solids are affected due to imperfactions.

Question 61.
Mention the types of defects in the solids or crystal structures.
Answer:
The defects in crystalline solids are of two types viz., (1) Point defect and (2) Line defect.
(1) Point defects are further classified as :
(a) Vacancy defect or Schottky defect
(b) Interstitial defect or Frenkel defect
(c) Impurity defect :

This is further classified as

• Substitution impurity defect
• Interstitial impurity defect.

(2) Line defects are further classified as

• Edge dislocation
• Screw dislocation.

Question 62.
What are point defects?
Answer:
Point defects : These defects arise due to irregularities produced in the arrangement of basis of lattice points in crystalline solids.

Question 63.
What are major classes of point defects ?
Answer:
There are three major classes of point defects : stoichiometric point defects, impurity defects and nonstoichiometric point defects.

There are four types of stoichiometric point defects as vacancy defect, self interstitial defect, Schottky defect and Frenkel defect.

Question 64.
Explain vacancy defect.
Answer:
Vacancy defect :
(1) During crystallisation, some of regular sites in solid remain unoccupied and the missing particle creates a vacancy defect.
(2) The defect can be developed by heating the substance.

(3) The mass of solid decreases due to absence of particles in regular sites.
(4) Since the volume remains the same the density of the substance decreases.

Question 65.
Explain self interstitial defect in elemental solid.
Answer:
Self Interstitial defect in elemental solid : The empty spaces or voids in between the particles at lattice points represent interstitial defective sites or self interstitial defects.

This defect arises in the following two ways :
(1) An extra particle occupies the empty interstitial space. This extra particle is similar to those already present in the crystal.

(2) (i) A particle gets shifted from its original regular site to an empty interstitial space in the crystal.

(ii) This displacement of a particle produces a vacancy defect at its regular site.
(iii) This defect is referred to as a combination of vacancy defect and self interstitial defect.
(iv) Since there is neither loss or gain in mass of the substance, the density of it remains unchanged.

Question 66.
How does Schottky defect arise?
Answer:
(1) Schottky defect arises in ionic solids due to missing of equal number of cations and anions from their regular positions in the crystal lattice creating vacancies.

(2) There arises formation of two holes per loss of ion pair.

(3) Conditions for formation of Schottky defect :
Characteristics of ionic solids showing Schottky defect :

• High degree of ionic character
• High coordination number of anion
• Small difference between ionic size or radii of cation and anion. The ionic ratio \(\frac{r_{\text {cation }}}{r_{\text {anion }}}\) is not below unity.

Question 67.
How does Frenkel defect arise?
Answer:

• Frenkel defect : This defect arises when an ion of an ionic compound is missing from its regular site and occupies interstitial vacant position between lattice points.
• Cations have smaller size than anions, hence generally cations occupy the interstitial sites.
• This creates a vacancy defect at its original position and interstitial defect at new position.
• Frenkel defect is regarded as the combination of interstitial defect and vacancy defect.

Conditions for the formation of Frenkel defect :

• This defect arises in ionic compounds with a large difference between the sizes of cation and anion.
• The ionic compounds must have ions with low coordination number.

Question 68.
What are the consequences of Frenkel defect ?
Answer:
Consequences of Frenkel defect :

• Since there is no loss of ions from the crystal lattice, the density of the solid remains unchanged.
• The crystal remains electrically neutral.
• This defect is observed in ZnS, AgCl, AgBr, AgI, CaF₂, etc.

Question 69.
Explain nonstoichiometric defect.
Answer:
Nonstoichiometric defect : This defect arises when the ratio of number of one kind of atoms to that of other kind of atoms (or ratio of number of cations to anions) becomes different from the actual stoichiometric formula. This involves the change in stoichiometry of the compound.

There are two types of nonstoichiometric defects as follows :
(1) Metal deficiency defect : This defect arises in compounds of metal which show variable oxidation states. In some metal crystals, positive metal ions are missing from their regular lattice sites. The extra negative charge is balanced by cations of the same metal with higher oxidation state than that of missing cation at site.

Consider a crystal of NiO. When one Ni²⁺ is missed from its lattice point, it creates a vacant site.

The deficiency of two positive charges is compensated by two Ni³⁺ ions at other lattice points of Ni²⁺ ions and the composition of NiO crystal becomes Ni_0.97O_1.0.

(2) Metal excess defect : There are two types of metal excess defect as follows :
(a) Presence of a neutral atom or an extra positive ion at interstitial position :

There are two types or ways of metal excess defects in ZnO. In the first case, Zn atom is present in the interstitial space as shown in figure.

(b) Metal ions and electrons at interstitial sites :
The second case arises when ZnO is heated, Zn²⁺ and electrons are obtained,

The excess Zn²⁺ ions are trapped in interstitial sites in the crystal lattice. Electrons occupy interstitial sites by diffusing into the interstitial sites.
In above both cases, the nonstoichiometric formula of ZnO is Zn_{1 + x} O_1.0

Question 70.
Explain defects due to anion vacancies.
OR
Explain colour of crystals or F centres.
Answer:

• The defect due to anion vacancies imparts colour to the colourless crystal.
• When a colourless crystal of NaCl is heated in the atmosphere of sodium vapour, the sodium atoms are deposited on the crystal surface.
• Due to diffusion of Cl^– ions to the crystal surface vacancies are created at their regular sites.
• These diffused Cl^– ions combine with Na atoms on the surface forming NaCl along with releasing electrons from sodium atoms. Na + Cl^– → NaCl + e^–

The released electrons diffuse into the crystal and occupy vacant sites of anions Cl^– in the crystal.
The anion vacant sites occupied by electrons are called F-centres or colour centres. Due to colour centres NaCl shows yellow colour.
Now NaCl crystal has excess of Na atoms having nonstoichiometric formula Na_1+x Cl_1.0.

Question 71.
How are solids classified according to electrical conductivity?
Answer:
According to electrical conductivity, solids are classified as follows :
(1) Conductors :

• The solids having electrical conductivity in the range of 10⁴ to 10⁷ Ohm^-1m^-1 are called conductors.
• The examples of conductors are metals and electrolytes.
• Electrical conductivity in metals is due to free movement of electrons while electrolytes conduct electricity due to migration ions.

(2) Insulators :

• Solids having very low electrical conductivity in the range of 10^-20 to 10^-10 Ohm^-1 m^-1 are called insulators.
• The examples of insulators are nonmetals and molecular solids.

(3) Semiconductors :

• Solids having conductivity in the range of 10^-6 to 10⁴ Ohm^-1m^-1 are semiconductors.
• The conductivity range is intermediate between conductors and insulators.
• The examples of semiconductors are silicon and germanium.

Question 72.
Explain band theory.
OR
Explain the origin of electrical properties in solids.
Answer:
(1) Metals are good conductors of heat and electricity. This is explained on the basis of band theory which involves the presence of free electrons.

(2) According to band theory, the atomic orbitals of metal atoms overlap to form molecular orbitals which are spread all over the crystal structure.

(3) The energy difference between adjacent molecular orbitals decreases as the number of molecular orbitals increases and when it becomes very less, the orbitals merge into one another forming continuous bands which extent over the entire crystal.

(4) There are two types of bands of molecular orbitals as follows :

• Valence band : The atomic orbitals with filled electrons from the inner shells form valence bands, where there are no free mobile electrons since they are involved in bonding.
• Conduction band : Atomic orbitals which are partially filled or empty on overlapping form closely placed molecular orbitals giving conduction bands where electrons are delocalised and can conduct, heat and electricity.

(5) Band gap :

• The energy difference between valence band and conduction band is called band gap.
• Band gap decides whether electrons from valence band can be promoted to vacant conduction band or not.
• The conductors like metals have very small or no band gap and electron can be promoted by thermal energy. The nonconductors have large band gap. The insulators have very large band gap.

Question 73.
Metals are good conductors of electricity. Explain.
Answer:
(1) Metals are good conductors of electricity since the outermost electrons of atoms in metallic crystal occupy conduction bands.

(2) The number of electrons in conduction bands is very large.
(3) The conduction bands in metals can be labelled as ‘s’ band, overlapping s and p bands etc. according to overlapping of orbitals.
(4) This results in delocalisation of outermost electrons forming metal ions. Hence, this is analogous to metal cations immersed in the sea of electrons.

Question 74.
Why does metallic conductivity decrease by increasing temperature?
Answer:

1. In metals a large number of outermost electrons of atoms occupy conduction bands.
2. Band formation in metals results in delocalisation of outermost electrons forming metal ions or cations.
3. The metallic cations occupying crystal lattice sites vibrate about mean positions.
4. As temperature increases the vibrational motion increases which interrupts flow of electrons decreasing electrical conductivity.

Question 75.
Explain insulators.
Answer:
(1) In insulators the valence band is completely filled by electronics while conduction band is empty.

(2) The energy gap between valence band and conduction band in insulator is very large.
(3) Thermal energy is not sufficient to promote electrons from valence band to conduction band.
(4) Therefore, the conduction band in insulator remains vacant and does not allow the conduction of electricity.

Question 76.
What are semiconductors ? Mention the types of semiconductors.
Answer:
(1) Semiconductors : The substances like silicon, germanium which have poor electrical conductance at low temperature but the conductance increases with the increase in temperature are called semiconductors.

(2) Their conductivity lies between metallic conductors and insulators.
(3) The energy difference between valence band and conduction band is relatively small, hence the electrons from valence band can be excited to conduction band by heating.

(4) Types of semiconductors : There are two types of semiconductors :
(a) Intrinsic semiconductor
(b) Extrinsic semiconductor

(a) Intrinsic semiconductor :

• A pure semiconductor material like pure Si, Ge which have a very low but finite electrical conductivity is called intrinsic semiconductor.
• The electrical conductivity of a semiconductor increases with the increase in temperature.

(b) Extrinsic semiconductor :

• Semiconductor doped with different element is called extrinsic semiconductor.
• By doping with elements like Ga or P, the electrical conductivity is increased.

Question 77.
Explain extrinsic semiconductor and doping.
Answer:
(1) A semiconductor obtained by doping intrinsic semiconductor with elements of third group and fifth group is called extrinsic semiconductor.
(2) This extrinsic semiconductor has higher electrical conductivity than pure intrinsic semiconductor.

(3) There ate two types of extrinsic semiconductors:
(A) n-type semiconductors:
(i) n-type semiconductor contains increased number of electrons in the conduction band.
(ii) When Si semiconductor is doped with 15th group element phosphorus, P, the new atoms occupy some vacant sites in the lattice in place of Si atoms.

(iii) P has five valence electrons, out of which four are involved in covalent bonding with neigh-bouring Si atoms while one electrons remains free and delocalised.
(iv) These free electrons increase the electrical conductivity of the semiconductor.
(v) The semiconductors with extra non-bonding free electrons are called n-type semiconductors.

(B) p-type semiconductor :
(i) p-type semiconductor is obtained by doping a pure semiconductor by an element of 13th group like B.
(ii) 13th group element has less number of valence electrons. When pure Si is doped with B atoms, these atoms occupy Si lattice points.

(iii) Boron (₅B) has only 3 valence electrons which form covalent bonds with the neighbouring Si atoms, while one bond has shortage of one electron.
(iv) This creates a vacancy or a hole, hence the electron from neighbouring Si atom jumps into this hole creating a vacancy in itself. This process continues, i.e., positive holes move in one direction while electrons moves in opposite direction.
(v) Due to electron deficient positions, this semiconductor is called p-type semiconductor.
(vi) When p-type semiconductor is connected to the external source of electricity, electrons from neighbouring silicon atoms jump into the holes so that electrons move towards positive electrode and holes migrate towards negative electrode.
(vii) Hence electrical conduction in p-type semiconductor is due to electrons and holes.

Question 78.
What are the uses of semiconductors ?
Answer:
The uses of semiconductors are as follows :

1. They are used in transistors, digital computers and cameras.
2. They are used in solar cells and television sets.
3. By combining n-type and p-type semiconductors, n-p junctions are formed which are effectively used in rectifiers or to convert light energy into electrical energy.

Question 79.
Classify the following semiconductors into n or p-type.
(i) B doped with Si
(ii) As doped with Si
(iii) P doped with Si
(vi) Ge doped with In.
Answer:

Question 80.
Explain the origin of magnetic properties in solids.
Answer:
(1) The magnetic properties of a substance arise due to the presence of electrons in their atoms or molecules.

(2) The electrons while revolving around the nucleus in various orbits, also spin around their own axes. Both these motions of electrons result in generating magnetic field and magnetic moments. Hence electron be haves as a tiny magnet.

(3) An atomic orbital can accommodate maximum two electrons with opposite spins, clockwise and anticlockwise. The degenerate orbitals like p, d and f orbitals can accommodate electrons with same spins until they are half filled.

(4) When a substance contains one or more unpaired electrons spinning in same direction, then their magnetic moments and magnetic properties add and the substance is said to be paramagnetic.

If a substance contains all electrons paired then their spins are balanced and magnetic moments and magnetic properties are cancelled and the substance is said to be diamagnetic.

Question 81.
Explain diamagnetism.
Answer:
(1) The magnetic properties of a substance arise due to presence of the electrons.

(2) An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic moment and a magnetic property.

(3) If an atom or a molecule of the substance contains all electrons paired, spinning clockwise and anticlockwise, their magnetic moments and magnetic properties get cancelled. Hence they oppose and repel the applied magnetic field. This phenomenon is called diamagnetism and the substance is said to be diamagnetic.
For example : Zn, Cd, H₂O, NaCl, etc.

Multiple Choice Questions

Question 82.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. A substance which on cutting will give irregular cleavage is
(a) glass
(b) KBr
(c) ZnS
(d) NaCl
Answer:
(a) glass

2. A solid which has definite heat of fusion is
(a) plastic
(b) CaCl₂
(c) glass
(d) soda lime glass
Answer:
(b) CaCl₂

3. In solids the constituent particles may be
(a) atoms
(b) ions
(c) molecules
(d) any one of the above three
Answer:
(d) any one of the above three

4. A single substance that exists in two or more forms is called
(a) polymorphous
(b) amorphous
(c) isomorphous
(d) monomorphous
Answer:
(a) polymorphous

5. Graphite is a
(a) metallic crystal
(b) covalent crystal
(c) ionic crystal
(d) molecular crystal
Answer:
(b) covalent crystal

6. Anisotropy is observed in
(a) Pyrex glass
(b) plastic
(c) K₂SO₄
(d) fullerene
Answer:
(c) K₂SO₄

7. The number of crystal systems (or types) is
(a) 4
(b) 7
(c) 8
(d) 12
Answer:
(b) 7

8. The number of Bravais lattices are
(a) 6
(b) 8
(c) 12
(d) 14
Answer:
(d) 14

9. Face centred cubic crystal is
(a) Cubic lattice of Bravais system
(b) Bravais lattice of HCP
(c) Bravais lattice of cubic system
(d) Cubic lattice with 5 atoms
Answer:
(c) Bravais lattice of cubic system

10. The number of tetrahedral sites per sphere in ccp structure is,
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

11. The packing fraction for a body centred cubic structure is
(a) 0.42
(b) 0.53
(c) 0.68
(d) 0.82
Answer:
(c) 0.68

12. If r is the radius of an atom and a is an edge length of fcc unit cell, then
(a) r = \(\frac{\sqrt{3}}{4} a\)
(b) r = \(\frac{a}{2 \sqrt{2}}\)
(c) r = \(\frac{a}{2}\)
(d) r = \(2 \sqrt{2} a\)
Answer:
(b) r = \(\frac{a}{2 \sqrt{2}}\)

13. The ratio of packing efficiency in see, bcc and fee crystalline structures is
(a) 1 : 1.2 : 1.3
(b) 1 : 1.12 : 1.23
(c) 1 : 1.3 : 1.4
(d) 1 : 1.25 : 1.38
Answer:
(c) 1 : 1.3 : 1.4

14. The correct sequence of the atomic layers in cubic close packing is
(a) ABABA
(b) ABACABAC
(c) ABCABC
(d) AABBAABB
Answer:
(c) ABCABC

15. The major binding force in diamond is
(a) Covalent bond
(b) Ionic bond
(c) Metallic bond
(d) Coordinate covalent bond
Answer:
(a) Covalent bond

16. The ratio of close packed atoms to octahedral holes in cubic packing is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(a) 1 : 1

17. A defect present in AgCl is
(a) Frenkel defect
(b) Schottky defect
(c) point defect
(d) interstitial impurity defect
Answer:
(a) Frenkel defect

18. An ionic solid crystallises in bcc structure. If the ionic radii of cation and anion are 0.84 Å and 1.07 Å, the length of the body diagonal is
(a) 1.91 Å
(b) 2.75 Å
(c) 3.82 Å
(d) 2.32 Å
Answer:
(c) 3.82 Å

19. The type of defect in NaCl crystal will be
(a) point defect
(b) interstitial defect
(c) vacancy defect
(d) impurity defect
Answer:
(c) vacancy defect

20. Schottky defects are observed in which solid among the following ?
(a) Brass
(b) Cesium chloride
(c) Zinc sulphide
(d) Stainless steel
Answer:
(b) Cesium chloride

21. An ionic compound crystallises in FCC type structure with ‘A’ ions at the centre of each face and ‘B’ ions occupying comers of the cube. The formula of compound is
(a) AB₄
(b) A₃B
(c) AB
(d) AB₃
Answer:
(b) A₃B

22. Total number of different primitive unit cells are
(a) 6
(b) 7
(c) 12
(d) 14.
Answer:
(d) 14

23. The volume of atoms present in body centred cubic unit cell of a metal of atomic radius r is,
(a) \(\frac{16}{3} \pi r^{3}\)
(b) \(\frac{8}{3} \pi r^{3}\)
(c) \(\frac{12}{3} \pi r^{3}\)
(d) \(\frac{24}{3} \pi r^{3}\)
Answer:
(b) \(\frac{8}{3} \pi r^{3}\)

24. The substances which can be permanently magnetised are
(a) diamagnetic
(b) paramagnetic
(c) ferromagnetic
(d) non-magnetic
Answer:
(c) ferromagnetic

25. CrO₂ is
(a) diamagnetic
(b) paramagnetic
(c) metalic
(d) ferromagnetic
Answer:
(d) ferromagnetic

26. A metallic element crystallises in face centred cubic structure. If the radius of metal ion is 0.92 Å, the edge length of the unit cell of the crystal is
(a) 0.8464 Å
(b) 1.252 Å
(c) 5.187 Å
(d) 2.6 Å
Answer:
(d) 2.6 Å

27. The volume of unit cell of a metallic crystal of bcc type is 8.4 × 10^-23 cm³. The volume occupied by 10 atoms in the crystalline structure is
(a) 4.2 × 10^-22 cm³
(b) 3.12 × 10^-23 cm³
(c) 1.74 × 10^-23 cm³
(d) 2.856 × 10^-22 cm³
Answer:
(d) 2.856 × 10^-22 cm³

28. Copper crystallises in face centred cubic structure. If the unit cell length is 360 pm, the radius of copper atom is
(a) 180 pm
(b) 156 pm
(c) 127 pm
(d) 110 pm
Answer:
(c) 127 pm

29. If all the lattice points in ccp structure namely comers, face and edge centres and body centre are occupied by atoms then the total number of atoms in the unit cell will be
(a) 8
(b) 12
(c) 14
(d) 16
Answer:
(a) 8

30. Gold crystallises in face centred cubic structure. If atomic mass of gold is 197 g mol^-1, the mass of the unit cell of gold will be
(a) 3.25 × 10^-23 kg
(b) 6.5 × 10^-23 kg
(c) 3.9 × 10^-24 kg
(d) 1.3 × 10^-24 kg
Answer:
(d) 1.3 × 10^-24 kg

31. The mass of a unit cell of a body centred cubic crystal of a metal is 72.2 × 10^-23 g. The atomic mass of the metal is
(a) 128.6 gmol^-1
(b) 108.7 gmol^-1
(c) 217.3 gmol^-1
(d) 57.86 gmol-1
Answer:
(c) 217.3 gmol^-1

32. An element crystallises in fee structure. If the atomic mass of the element is 72.7 U, the mass of one unit cell of it will be
(a) 2.9 × 10^-24 g
(b) 4.83 × 10^-25 kg
(c) 1.2 × 10^-22 g
(d) 2.41 × 10^-24 kg
Answer:
(b) 4.83 × 10^-25 kg

33. Edge length of a cubic unit cell is 354 pm. The distance between two atoms diagonally opposite on the face is
(a) 500 pm
(b) 354 pm
(c) 708 pm
(d) 627 pm
Answer:
(a) 500 pm

34. The unit cell has an edge length 403 pm. The distance between two atoms placed opposite ends of body diagonal will be
(a) 806 pm
(b) 201.5 pm
(c) 698 pm
(d) 578 pm
Answer:
(c) 698 pm

35. An element having atomic mass 115 u, crystallises in bcc structure. The number of unit cells in 1 g of the element will be
(a) 2.6 × 10²¹
(b) 3.8 × 10²³
(c) 8.7 × 10^-3
(d) 6.17 × 10²⁰
Answer:
(a) 2.6 × 10²¹

36. The edge length of a bcc unit cell of a metallic crystal is 2.9 Å. Hence the diameter of an atom is
(a) 1.025 Å
(b) 2.512 Å
(c) 1.45 Å
(d) 1.31 Å
Answer:
(b) 2.512 Å

37. An element crystallises in fee structure. If the atomic radius is 130 pm, the edge length of unit cell is
(a) 332.5 pm
(b) 410 pm
(c) 390 pm
(d) 367.6 pm
Answer:
(d) 367.6 pm

38. The arrangement of layers in hexagonal close packing is
(a) ABCABC
(b) ABAB
(c) ABBABBA
(d) ABBCABBC
Answer:
(b) ABAB

39. For square close packing, the planar arrangement is
(a) AAAA
(b) ABAB
(c) ABCABC
(d) AABBAA
Answer:
(a) AAAA

40. Semiconductors are manufactured by addition of impurities of
(a) p-block elements
(b) actinoids
(c) Lanthanoids
(d) s-block elements
Answer:
(a) p-block elements

41. p-type semiconductor is formed when trace amount of impurity is added to silicon. The number of valence electrons in the impurity atom must be
(a) 3
(b) 5
(c) 1
(d) 2
Answer:
(a) 3

42. n-type semiconductor is formed when trace amount of impurity is added to silicon. The number of valence electrons in the impurity atom must be
(a) 3
(b) 5
(c) 1
(d) 2
Answer:
(b) 5