CBSE Class 12 Chemistry Solutions MCQs Set 18

Answer: (a) K kg mol⁻¹

Explanation:
1. K_b = \( \frac{\Delta T_b}{m} \) = \( \frac{K}{mol\ kg^{-1}} \) = K kg mol⁻¹

Teacher's Note:
1. Derive units from ΔT_b = K_bm rather than memorising them.

Answer: (a) 373.2 K

Explanation:
1. Molality = \( \frac{18/180}{1\ \text{kg}} \) = 0.1 m; ΔT_b = 0.52 × 0.1 = 0.052 K
2. T_b = 373.15 + 0.052 = 373.202 K ≈ 373.2 K

Teacher's Note:
1. Elevation for dilute solutions is small (a fraction of a degree); large answers usually signal a unit slip.

Answer: (a) freezing point is raised

Explanation:
1. KI reacts with HgI₂ to form a complex as
2. 2KI + HgI₂ → K₂[HgI₄] → 2K⁺ + [HgI₄]²⁻; (2K⁺ + 2I⁻) = 4 ions become 3 ions
3. Thus, number of ions decreases in the solution. Hence, depression in freezing point is less, i.e., actual freezing point is raised.

Teacher's Note:
1. Count the particles before and after the reaction: fewer particles means a smaller depression, so a higher freezing point.

Answer: (d) unchanged

Explanation:
1. Molal depression constant (K_f) depends only on the solvent and is independent of the molality of the solution.

Teacher's Note:
1. Doubling molality doubles ΔT_f, not K_f.

Answer: (b) 0.1 M Na₃PO₄

Explanation:
1. Higher the concentration of ions in the solution, higher is the elevation in boiling point. As 0.1 M Na₃PO₄ has highest concentration of ions, elevation of boiling point is highest for 0.1 M Na₃PO₄ and so is the boiling point.
2. Alternatively, ΔT_b = i K_b m. For the same value of m, higher is the value of i, higher is ΔT_b.

Teacher's Note:
1. Compare i values: KNO₃ 2, BaCl₂ 3, K₂SO₄ 3, Na₃PO₄ 4.

Answer: (c) three times

Explanation:
1. MgCl₂ → Mg²⁺ + 2Cl⁻, so i = 3; glucose does not ionise, i = 1.
2. ΔT_f ∝ i, so the depression is three times.

Teacher's Note:
1. Assume complete dissociation of a strong electrolyte unless the question gives a degree of dissociation.

Answer: (c) solution B is greater than A

Explanation:
1. 7 g L⁻¹ MgCl₂ = \( \frac{7}{24 + 71} \) mol L⁻¹ = \( \frac{7}{95} \) mol L⁻¹ = \( \frac{7 \times 3}{95} \) mol L⁻¹ of ions = 0.22 M
2. 7 g L⁻¹ NaCl = \( \frac{7}{23 + 35.5} \) M = \( \frac{7}{58.5} \) M = \( \frac{7 \times 2}{58.5} \) mol L⁻¹ of ions = 0.24 M
3. As concentration of ions in NaCl solution is greater, NaCl solution (solution B) will have greater osmotic pressure.

Teacher's Note:
1. Equal masses do not mean equal particles: convert to moles of ions before comparing.

Answer: (c) x < z < y

Explanation:
1. For association, i < 1. For dissociation, i > 1.
2. For no change, i = 1. Hence, order is x < z < y.

Teacher's Note:
1. Association reduces the number of particles (i < 1); dissociation increases it (i > 1).

Answer: (d) K₄[Fe(CN)₆]

Explanation:
1. Al₂(SO₄)₃ and K₄[Fe(CN)₆] produce the same number of ions in the solution (viz. 5). Thus, they have the same value of van't Hoff factor.

Teacher's Note:
1. A complex ion such as [Fe(CN)₆]⁴⁻ counts as one particle.

Answer: (c) Al₂(SO₄)₃

Explanation:
1. ΔT_f = i × K_f × m. As K_f and m are same for all given solutions, ΔT_f ∝ i
2. and i for KCl = 2; for C₆H₁₂O₆, i = 1; for Al₂(SO₄)₃, i = 5 and for K₂SO₄, i = 3.
3. Hence, Al₂(SO₄)₃ will show largest freezing point depression.

Teacher's Note:
1. Largest depression means lowest freezing point – both phrasings point to the largest i.

Answer: (b) 60

Explanation:
1. i = \( \frac{M_c}{M_o} \) = \( \frac{40}{25} \) = 1.6
2. NaOH → Na⁺ + OH⁻; 1 mol → (1 − α), α, α; Total = 1 + α
3. i = 1 + α or α = i − 1 = 0.6 = 60%

Teacher's Note:
1. van't Hoff factor = calculated (normal) molar mass ÷ observed molar mass; dissociation makes the observed value smaller.

Answer: (d) 0.04

Explanation:
1. CH₃COOH ⇌ CH₃COO⁻ + H⁺; 1 mole → 1 − α, α, α; Total = 1 + α
2. i = 1 + α or α = i − 1 = 1.04 − 1 = 0.04

Teacher's Note:
1. For a binary electrolyte (AB → A⁺ + B⁻), α = i − 1; the volume data is not needed.

Answer: (c) X is undergoing dissociation in water

Explanation:
1. For equimolal solutions of two substances, the one that undergoes dissociation produces greater number of particles (ions) in the solution and hence has greater boiling point.

Teacher's Note:
1. For equimolal solutions molar mass is irrelevant; only i decides ΔT_b.

Answer: (a) 49

Explanation:
1. BaCl₂ ⇌ Ba²⁺ + 2Cl⁻; Initial 1 mole, 0, 0; After disso. 1 − α, α, 2α
2. Total no. of particles = 1 + 2α; i = 1 + 2α
3. or α = \( \frac{i - 1}{2} \) = \( \frac{1.98 - 1}{2} \) = \( \frac{0.98}{2} \) = 0.49 = 49%

Teacher's Note:
1. For an electrolyte giving n ions, α = (i − 1)/(n − 1); here n = 3.

Answer: (d) 3

Explanation:
1. As Ba(OH)₂ is a strong electrolyte, it dissociates completely as Ba(OH)₂ → Ba²⁺ + 2OH⁻
2. Ions produced = 3. Hence, van't Hoff factor = 3

Teacher's Note:
1. For a strong electrolyte in dilute solution, i equals the number of ions per formula unit.

Answer: (c) Ethanol and acetone

Explanation:
1. In ethanol, molecules are held by hydrogen bonds. Acetone molecules come in between and break some of these bonds, so A–B forces are weaker; the solution shows positive deviation and is non-ideal.
2. The other pairs have similar structures and similar intermolecular forces, so they form nearly ideal solutions.

Teacher's Note:
1. Look for a hydrogen-bonded liquid mixed with a non-H-bonding one – that pair deviates from ideality.

Answer: (a) shows a negative deviation from Raoult's law

Explanation:
1. In negative deviation the vapour pressure is lower than expected, so at a particular composition the vapour pressure is minimum and boiling point is maximum (maximum boiling azeotrope).

Teacher's Note:
1. Negative deviation → minimum vapour pressure → maximum boiling azeotrope (e.g., HNO₃–water).

Answer: (d) Solids and liquids are highly incompressible

Explanation:
1. Solids and liquids are highly incompressible, so a change in pressure hardly affects the solubility of a solid in a liquid.

Teacher's Note:
1. Option (c) describes Henry's law, which applies to gases, not solids.

Answer: (c) \( \frac{1}{76} \)

Explanation:
1. \( \frac{p^\circ - p_s}{p^\circ} \) = \( \frac{n_2}{n_1 + n_2} \) = x₂, i.e., \( \frac{760 - 750}{760} \) = x₂
2. or x₂ = \( \frac{10}{760} \) = \( \frac{1}{76} \)

Teacher's Note:
1. At 373 K the vapour pressure of pure water is 760 mm Hg – the question expects you to know this.

Answer: (a) − 0.372°C

Explanation:
1. ΔT_f = K_f × m = 1.86 × 0.2 = 0.372
2. ∴ T_f = 0 − 0.372 = − 0.372°C

Teacher's Note:
1. ΔT_f is a depression: subtract it from 0°C, so the answer must be negative.

Answer: (c) Decrease in Molarity

Explanation:
1. On increasing the temperature of an aqueous solution, volume of the solution increases but number of moles of the solute dissolved remains the same. Hence, molarity decreases.

Teacher's Note:
1. Molality (mass based) does not change with temperature; molarity (volume based) does.

Answer: (a) ΔH_mix = 0 and ΔV_mix = 0

Explanation:
1. For an ideal solution, there is no heat change and no volume change on mixing: ΔH_mix = 0 and ΔV_mix = 0.

Teacher's Note:
1. (b) describes positive deviation and (c) negative deviation – know all three sets.

Answer: (c) Osmotic pressure

Explanation:
1. Osmotic pressure is measured at room temperature and has a measurable magnitude even for very dilute solutions, so it is used for polymers and proteins (which are also often unstable at high temperatures).

Teacher's Note:
1. For macromolecules the other colligative effects are too small to measure accurately.

Answer: (d) Both of them to distil over in the same composition as that of the mixture being distilled

Explanation:
1. On distilling an azeotropic mixture, both the liquids distil over in the same composition.

Teacher's Note:
1. An azeotrope boils at constant temperature with vapour of the same composition as the liquid, so it cannot be separated by fractional distillation.

Answer: (a) 36 g mol⁻¹

Explanation:
1. For isotonic solutions, Molar concentration of glucose = Molar concentration of X
2. Molar conc. of glucose = \( \frac{5}{180} \) × \( \frac{1}{100} \) × 1000 = \( \frac{5}{18} \) mol L⁻¹
3. Molar conc. of X = \( \frac{1}{M_X} \) × \( \frac{1}{100} \) × 1000 = \( \frac{10}{M_X} \) mol L⁻¹
4. ∴ \( \frac{10}{M_X} \) = \( \frac{5}{18} \) or M_X = 36 g mol⁻¹

Teacher's Note:
1. Isotonic solutions have equal molar concentrations of particles (at the same temperature).

Answer: (b) 50 g mol⁻¹

Explanation:
1. M₂ = \( \frac{1000 \times K_b \times w_2}{\Delta T_b \times w_1} \) = \( \frac{1000 \times 0.52 \times 2.5}{0.52 \times 50} \)
2. = 50 g mol⁻¹ (50 mL of water = 50 g of water)

Teacher's Note:
1. Take the density of water as 1 g mL⁻¹ to convert the volume of water to mass.

Answer: (d) High pressure

Explanation:
1. By Henry's law, the solubility of a gas increases with its partial pressure, so soft drink bottles are sealed under high pressure of CO₂.

Teacher's Note:
1. This is the standard Henry's law application; fizz on opening shows the drop in solubility when pressure falls.

Answer: (b) The intermolecular forces A – A, B – B < A – B

Explanation:
1. In negative deviation, A–B attractions are stronger than A–A and B–B attractions, so molecules escape less easily and p < p°x.
2. Options (a) and (c) describe positive deviation, so (d) is also wrong.

Teacher's Note:
1. Negative deviation: stronger A–B forces, lower vapour pressure, ΔH_mix < 0 and ΔV_mix < 0.

Answer: (d) π = CRT : Osmotic pressure :: P > π : Reverse osmosis

Explanation:
1. π = CRT is the expression for osmotic pressure, and applying a pressure P greater than π on the solution side causes reverse osmosis.
2. In (a) the deviations are interchanged; (b) and (c) mismatch the laws/solution types.

Teacher's Note:
1. In analogy questions check both halves – one wrong half makes the whole option wrong.

Answer: (c) n-hexane and n-heptane

Explanation:
1. n-Hexane and n-heptane are similar non-polar molecules with nearly equal intermolecular forces, so they form an ideal solution.
2. Chloroform–acetone and phenol–aniline show negative deviation; ethanol–acetone shows positive deviation.

Teacher's Note:
1. Members of the same homologous series usually form nearly ideal solutions.

Answer: (c) P_Total = P_Solvent

Explanation:
1. As non-volatile solute does not form vapour, total vapour pressure of solution will be due to solvent in the solution.

Teacher's Note:
1. With a non-volatile solute, P_total = p_solvent = p°_solvent x_solvent.

Answer: (a) < 50 mL

Explanation:
1. Acetone-chloroform mixture shows negative deviation from Raoult's law therefore, ΔV_mixing = −ve, i.e., V_total < 50 mL.

Teacher's Note:
1. Acetone and chloroform hydrogen-bond with each other, pulling molecules closer, so volume contracts.

Answer: (a) 0.08

Explanation:
1. By Henry's law, x = p/K_H, i.e., x ∝ p at constant temperature.
2. Doubling the pressure doubles the mole fraction: 2 × 0.04 = 0.08

Teacher's Note:
1. Henry's law is a direct proportion at constant temperature – no need to find K_H.

Answer: (d) Freezing point of solution

Explanation:
1. As vapour pressure of solvent is higher than that of solution, the upper curve is for the solvent. The point X lies in the solid solvent curve. The lower solution curve meets the solid solvent curve at X. Hence, X represents the freezing point of the solution.

Teacher's Note:
1. Freezing point is where the vapour pressure of the liquid (solution) equals that of the solid solvent.

Answer: (b) Raoult's law

Explanation:
1. Raoult's law: p_A = p_A° x_A for each volatile component of a solution.

Teacher's Note:
1. Raoult's law uses the pure-component vapour pressure as the constant; Henry's law uses K_H for a gas.

Answer: (b) exothermic and reversible process

Explanation:
1. Gas + Liquid ⇌ Solution; ΔH < 0. Dissolution of a gas is exothermic and reversible, so by Le Chatelier's principle, raising temperature shifts the equilibrium backward and solubility decreases.

Teacher's Note:
1. Only an equilibrium (reversible) process responds to temperature through Le Chatelier's principle.

Answer: (a) 7.5 mol

Explanation:
1. ΔT_f = \( \frac{1000 \times K_f \times w_2}{M_2 \times w_1} \)
2. 14 = \( \frac{1000 \times 1.86 \times w_2}{M_2 \times 1000} \) = \( \frac{1000 \times 1.86 \times n_2}{1000} \) \( \left(\because \frac{w_2}{M_2} = n_2\right) \)
3. or n₂ = \( \frac{14}{1.86} \) = 7.5 mol

Teacher's Note:
1. 1 litre of water is taken as 1 kg, so the molality equals the moles of alcohol added.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Inside a pressure cooker the pressure is higher than atmospheric, so water boils at a higher temperature.
2. Food cooks faster at this higher temperature, so R correctly explains A.

Teacher's Note:
1. A liquid boils when its vapour pressure equals the external pressure; higher external pressure means a higher boiling point.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. The boiling point of the azeotropic mixture may be greater or less depending upon whether solution shows negative or positive deviation from Raoult's law.

Teacher's Note:
1. Positive deviation gives a minimum-boiling azeotrope; negative deviation gives a maximum-boiling azeotrope.

Answer: (c) A is true, but R is false.

Explanation:
1. Ice is less dense than water, so increasing pressure favours melting and lowers the freezing point; A is true.
2. Correct R. Density of water is maximum at 277 K (4°C).

Teacher's Note:
1. Remember 4°C (277 K) for the maximum density of water, not 0°C.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. ΔT_b = i K_b m, ΔT_f = i K_f m. Although i is same but K_f and K_b are not equal and hence ΔT_b ≠ ΔT_f.

Teacher's Note:
1. For water K_f = 1.86 and K_b = 0.52 K kg mol⁻¹, so ΔT_f is about 3.6 times ΔT_b.

Answer: (c) A is true, but R is false.

Explanation:
1. Benzoic acid dimerises in benzene through hydrogen bonding, so the number of particles decreases and i < 1; A is true.
2. Correct R. Benzoic acid behaves as a weak electrolyte in water.

Teacher's Note:
1. In benzene, carboxylic acids associate (dimers), giving i close to 0.5.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. Relative lowering of vapour pressure depends upon the mole fraction of the solute in the solution.

Teacher's Note:
1. (p° − p_s)/p° = x₂, the mole fraction of the solute.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. Through the process of osmosis, raw mangoes shrivel into pickle.

Teacher's Note:
1. No external pressure is applied, so water leaves the mango by ordinary osmosis.

Answer: (c) A is true, but R is false.

Explanation:
1. Molarity depends on the volume of solution, which changes with temperature; A is true.
2. Correct R. Molarity is not a colligative property.

Teacher's Note:
1. Molarity is a unit of concentration; colligative properties are ΔT_b, ΔT_f, π and relative lowering of vapour pressure.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. Cryoscopic constant is not a universal constant. Its value depends upon the solvent.

Teacher's Note:
1. K_f is different for every solvent (water 1.86, benzene 5.12, camphor about 40 K kg mol⁻¹).

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. BaCl₂ dissociates to produce 3 ions (Ba²⁺ + 2Cl⁻) whereas NaCl dissociates to give 2 ions (Na⁺ + Cl⁻). Hence, elevation in boiling point of BaCl₂ solution will be 1.5 times that of NaCl.
2. Ba²⁺ does carry double the charge of Na⁺, so R is true (but charge does not affect colligative properties).

Teacher's Note:
1. Colligative properties depend on the number of ions, not their charge: compare i = 3 with i = 2.