CBSE Class 12 Chemistry Solutions MCQs Set 16

Answer: (c) Solute: Liquid; Solvent: Solid

Explanation:
1. Amalgam of mercury with sodium is classified as a solid solution in which the solute (mercury) is a liquid and the solvent (sodium) is a solid.

Teacher's Note:
1. Classify a solution by the physical state of the solvent: a solid solvent gives a solid solution, whatever the state of the solute.

Answer: (c) 6

Explanation:
1. ppm = \( \frac{\text{mass of O}_2}{\text{mass of solution}} \) × 10⁶ = \( \frac{6\times10^{-3}\text{ g}}{1000\text{ g}} \) × 10⁶ = 6 ppm

Teacher's Note:
1. 1 mg per kg is exactly 1 ppm by mass – a quick check for such questions.

Answer: (b) 0.0177

Explanation:
1. 1 mol solute in 1000 g water; moles of water = 1000/18 = 55.55 mol
2. x_solute = \( \frac{1}{1 + 55.55} \) = 0.0177

Teacher's Note:
1. Convert 1 kg of water to moles (55.55 mol) before applying the mole fraction formula.

Answer: (a) 35% volume by volume

Explanation:
1. A 35% (v/v) solution of ethylene glycol in water is used as an antifreeze (coolant) in car engines.

Teacher's Note:
1. Both components are liquids, so the natural way to express this concentration is volume percentage.

Answer: (d) Powdered sugar in hot water.

Explanation:
1. Dissolution is endothermic (solution feels cool), so it is favoured by higher temperature.
2. Powdered sugar has a larger surface area, so it dissolves faster. Hence powdered sugar in hot water dissolves most rapidly.

Teacher's Note:
1. Two factors act together here: an endothermic process speeds up on heating, and a larger surface area speeds up any dissolution.

Answer: (c) equal to the rate of crystallisation

Explanation:
1. In a saturated solution, a dynamic equilibrium exists: solute + solvent ⇄ solution, so the rate of dissolution equals the rate of crystallisation.

Teacher's Note:
1. Equilibrium is dynamic – both processes continue at equal rates, so the rate is not zero.

Answer: (d) both (a) and (b)

Explanation:
1. Solubility of a gas depends on the nature of the gas, the temperature and the pressure.
2. So at constant T, nature and pressure affect it; at constant P, nature and temperature affect it. Both (a) and (b) are correct.

Teacher's Note:
1. (c) is incomplete because it ignores the nature of the gas – different gases have different K_H values.

Answer: (b) greater for gases with lower solubility.

Explanation:
1. From p = K_Hx, at a given pressure x = p/K_H; so a larger K_H means a lower solubility.

Teacher's Note:
1. K_H is in the denominator of the solubility expression, so it varies inversely with solubility.

Answer: (b) low atmospheric pressure

Explanation:
1. At high altitude the partial pressure of oxygen is low; by Henry’s law less oxygen dissolves in blood, causing anoxia.

Teacher's Note:
1. Low temperature would actually increase gas solubility, so it cannot be the cause.

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

Explanation:
1. Chloroform forms a hydrogen bond with the oxygen of diethyl ether (Cl₃C–H···O(C₂H₅)₂).
2. A–B attractions are stronger than A–A and B–B, so the vapour pressure is lowered: negative deviation.

Teacher's Note:
1. New H-bonds formed on mixing mean stronger A–B interactions, which always point to a negative deviation.

Answer: (a) Methanol and acetone.

Explanation:
1. On adding acetone, the hydrogen bonds between methanol molecules are broken and the new interactions are weaker, so the mixture shows positive deviation.
2. The other three pairs form stronger interactions (H-bonds) on mixing and show negative deviation.

Teacher's Note:
1. Ask whether mixing breaks H-bonds (positive deviation) or creates new, stronger ones (negative deviation).

Answer: (c) (iii) and (iv)

Explanation:
1. Benzene and toluene form a nearly ideal solution, so Δ_mixH = 0 and Δ_mixV = 0 are true.
2. An ideal solution does not form any azeotrope, so statements (iii) and (iv) are not true.

Teacher's Note:
1. Azeotropes are formed only by solutions showing large deviations from Raoult’s law, never by ideal solutions.

Answer: (c) acetone-chloroform

Explanation:
1. Acetone–chloroform shows negative deviation (H-bonding between them), so it forms a maximum boiling azeotrope.
2. Ethanol–water and CS₂–acetone show positive deviation (minimum boiling azeotropes); benzene–toluene is nearly ideal.

Teacher's Note:
1. Negative deviation → maximum boiling azeotrope; positive deviation → minimum boiling azeotrope.

Answer: (a) is higher than that of a dilute solution.

Explanation:
1. π = CRT; at a given temperature π is directly proportional to the concentration, so a concentrated solution has a higher osmotic pressure.

Teacher's Note:
1. Like all colligative properties, osmotic pressure increases with the concentration of solute particles.

Answer: (d) it loses water due to osmosis.

Explanation:
1. The salt solution is hypertonic, so water moves out of the mango cells into the solution by osmosis and the mango shrivels.

Teacher's Note:
1. Reverse osmosis needs an external pressure greater than the osmotic pressure; nothing like that is applied here.

Answer: (a) the solution is hypotonic

Explanation:
1. Blood cells are isotonic with 0.9% (m/V) NaCl. A solution with less than 0.9% NaCl is hypotonic, so water enters the cells and they swell.

Teacher's Note:
1. Remember 0.9% (m/V) NaCl (normal saline) is isotonic with blood; below it is hypotonic, above it is hypertonic.

Answer: (d) Osmotic pressure

Explanation:
1. Osmotic pressure is measured at room temperature and has a large magnitude even for very dilute solutions, so it suits biomolecules that are unstable at high temperatures.

Teacher's Note:
1. Biomolecules are heat-sensitive and have high molar masses, which gives tiny ΔT values – that is why osmotic pressure is preferred.

Answer: (b) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region of lower concentration of solute to a region of higher concentration.

Explanation:
1. In reverse osmosis, solvent molecules move from the region of higher concentration of solute to the region of lower concentration under applied pressure. So statement (b) is false.

Teacher's Note:
1. Reverse osmosis is the opposite of osmosis: solvent is forced out of the concentrated solution.

Answer: (b) 2, 2 and 3

Explanation:
1. KCl → K⁺ + Cl⁻ (i = 2); NaCl → Na⁺ + Cl⁻ (i = 2); K₂SO₄ → 2K⁺ + SO₄²⁻ (i = 3)

Teacher's Note:
1. For complete dissociation, i equals the number of ions produced by one formula unit.

Answer: (c) 0.5

Explanation:
1. Ethanoic acid dimerises in benzene through hydrogen bonding: 2CH₃COOH ⇄ (CH₃COOH)₂
2. For complete dimerisation, i = 1/2 = 0.5

Teacher's Note:
1. Association gives i < 1 and dissociation gives i > 1.

Answer: (b) 1.0 M Na₂SO₄

Explanation:
1. Na₂SO₄ gives 3 ions (i = 3) while NaOH, NH₄NO₃ and KNO₃ give 2 ions each.
2. Since ΔT_b = iK_bm, Na₂SO₄ solution has the highest boiling point.

Teacher's Note:
1. At equal concentration, compare the number of particles each solute gives in solution.

Answer: (c) about three times

Explanation:
1. MgCl₂ → Mg²⁺ + 2Cl⁻, so i ≈ 3 while glucose does not dissociate (i = 1).
2. Hence ΔT_f of MgCl₂ solution is about three times that of glucose solution.

Teacher's Note:
1. Count the ions: one Mg²⁺ and two Cl⁻ make three particles.

Answer: (a) \( K_b = \frac{M_A R T_0^2}{1000\,\Delta_{vap}H} \)

Explanation:
1. From thermodynamics, \( K_b = \frac{R\,M_A\,T_0^2}{1000\,\Delta_{vap}H} \), where M_A is the molar mass of the solvent in g mol⁻¹.
2. Check: with R in J K⁻¹ mol⁻¹ and Δ_vapH in J mol⁻¹, the units come out as K kg mol⁻¹.

Teacher's Note:
1. For water, (8.314 × 18 × 373²)/(1000 × 40660) ≈ 0.51 K kg mol⁻¹ – a handy check of the formula.

Answer: (a) Two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point.

Explanation:
1. ΔT_f = K_fm, and K_f depends on the nature of the solvent. So solutions of the same molality in different solvents have different ΔT_f; statement (a) is false.

Teacher's Note:
1. Same molality gives the same ΔT_f only in the same solvent, because K_f is a property of the solvent.

Answer: (a) Mole fraction

Explanation:
1. According to Raoult’s law, the partial vapour pressure of a component is directly proportional to its mole fraction in the solution (p = p°x).
2. Hence, mole fraction is the unit that relates concentration with vapour pressure.

Teacher's Note:
1. Raoult’s law is written in mole fractions, so whenever vapour pressure is involved, convert the given concentration into mole fraction first.

Answer: (a) K kg mol⁻¹ or K (molality)⁻¹

Explanation:
1. It is the unit of ebullioscopic constant (K_b).
2. K_b = \( \frac{\Delta T_b}{m} \), so its unit is K kg mol⁻¹ or K (molality)⁻¹.

Teacher's Note:
1. Derive the unit from ΔT_b = K_bm: kelvin divided by mol kg⁻¹ gives K kg mol⁻¹. The same unit applies to K_f.

Answer: (c) pressure

Explanation:
1. The solubility of a solid in liquid does not depend upon pressure because solid is incompressible practically.

Teacher's Note:
1. Pressure matters a lot for gases dissolving in liquids (Henry’s law) but hardly at all for solids and liquids, which cannot be compressed.

Answer: (b) supersaturated

Explanation:
1. When a small amount of solute is added to its solution, it does not dissolve and gets precipitated then, this type of solution is called as supersaturated solution.

Teacher's Note:
1. In a supersaturated solution, adding a small crystal of solute makes the excess solute crystallise out; in a merely saturated solution the added solute just stays undissolved.

Answer: (d) 0.016

Explanation:
1. M = \( \frac{n}{V} \) = 0.02 = \( \frac{n}{4} \), so n = 0.08 mol
2. m = \( \frac{n}{\text{Mass of water in kg}} \) = \( \frac{0.08}{5} \) = 0.016 m

Teacher's Note:
1. Dilution does not change the number of moles of solute; only the amount of solvent changes. Here the dilute solution’s 5 L is taken as about 5 kg of water.

Answer: (b) Solution (A) will follow Raoult’s law.

Explanation:
1. Solution A will follow Raoult’s law due to nearly same interactions between A-A, B-B, and A-B. The solution formed will be nearly ideal.

Teacher's Note:
1. Link the interactions to the deviation: A–B weaker ⇒ positive deviation (B); A–B stronger ⇒ negative deviation (C); all similar ⇒ ideal (A).

Answer: (b) the number of solute particles in solution.

Explanation:
1. Colligative properties depend upon the number of solute particles in the solution and it is independent of its nature.

Teacher's Note:
1. The word ‘colligative’ means ‘bound together’: the property depends on how many particles are collected, not what they are.

Answer: (c) i_A = i_B = i_C

Explanation:
1. The value of van’t Hoff’s factor will be i. It is due to complete dissociation of strong electrolyte (NaCl) in dilute solutions and on complete dissociation value of i for NaCl is 2.

Teacher's Note:
1. This answer assumes complete dissociation of NaCl at all three concentrations (i = 2 each); in real solutions i creeps up towards 2 on dilution.

Answer: (b) 18 M

Explanation:
1. Molarity = \( \frac{\text{Mass \%} \times d \times 10}{M_B} \) = \( \frac{98 \times 1.8 \times 10}{98} \) = 18 M

Teacher's Note:
1. Mass % with density gives molarity directly; molality would need the mass of water (only 2 g per 100 g here), which gives a very different value.

Answer: (a) benzene + chloroform

Explanation:
1. Chloroform forms a weak C–H···π attraction with benzene, so the A–B interactions are not weaker than A–A and B–B; the mixture does not show positive deviation.
2. In the other three mixtures the solute–solvent attractions are weaker, so they show positive deviation.

Teacher's Note:
1. Positive deviation means A–B attractions are weaker than A–A and B–B, so molecules escape more easily and vapour pressure is higher than predicted.

Answer: (d) 0.1 M Na₃PO₄

Explanation:
1. Lowering of vapour pressure depends on the number of solute particles.
2. Na₃PO₄ → 3Na⁺ + PO₄³⁻ gives i = 4, the largest here, so its vapour pressure is the least.

Teacher's Note:
1. At equal molarity, compare van’t Hoff factors: urea 1, BaCl₂ 3, Na₂SO₄ 3, Na₃PO₄ 4.

Answer: (d) ΔS_mix = 0

Explanation:
1. For an ideal solution ΔH_mix = 0 and ΔV_mix = 0, but mixing always increases randomness, so ΔS_mix > 0.

Teacher's Note:
1. Mixing is spontaneous even for ideal solutions because ΔS_mix is positive, making ΔG_mix negative.

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

Explanation:
1. Molarity = moles of solute per litre of solution. Volume changes with temperature, so molarity changes; R correctly explains A.

Teacher's Note:
1. Molality and mole fraction, which use only masses, do not change with temperature.

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

Explanation:
1. R is Henry’s law (p = K_Hx, so x ∝ p), which directly explains why solubility increases with pressure.

Teacher's Note:
1. Henry’s law uses the partial pressure of the gas above the solution, not the total pressure.

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

Explanation:
1. Raoult’s law becomes a special case of Henry’s law when K_H = p°, so A is true.
2. R is also true (x = p/K_H), but it does not explain why Raoult’s law is a special case of Henry’s law.

Teacher's Note:
1. Both laws have the form p = constant × x; only the constant differs (p° vs K_H).

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

Explanation:
1. Only non-ideal solutions showing large deviations from Raoult’s law form azeotropes, so A is false.
2. R is true: minimum boiling azeotropes boil below and maximum boiling azeotropes boil above both components.

Teacher's Note:
1. Note the word “always” – such absolute words often make an assertion false.

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

Explanation:
1. The added liquid is more volatile and contributes its own partial vapour pressure, so the total vapour pressure exceeds that of the pure solvent; A is true.
2. When both components are volatile, both contribute to the vapour pressure, so R is false.

Teacher's Note:
1. R holds only when the solute is non-volatile.

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

Explanation:
1. A is true: solute particles occupy part of the surface and reduce the escaping tendency of solvent molecules.
2. R (Raoult’s law for relative lowering) is also true, but it gives the magnitude of lowering rather than the reason for it.

Teacher's Note:
1. The cause of the lowering is reduced surface area for evaporation; the formula only tells you how much it is lowered.

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

Explanation:
1. Adding a non-volatile solute lowers the vapour pressure, so the solution must be heated to a higher temperature to boil. The boiling point of the solution is higher than that of the pure solvent; A is false.
2. R is true.

Teacher's Note:
1. Lower vapour pressure means a higher boiling point – this is the elevation of boiling point.

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

Explanation:
1. The solution has a lower vapour pressure than the solvent, so the vapour pressures of solid and liquid become equal at a lower temperature. Hence the freezing point is depressed; R explains A.

Teacher's Note:
1. Freezing point is the temperature at which the vapour pressure of the liquid equals that of the solid.

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

Explanation:
1. A correctly describes osmosis.
2. R is false: solvent moves from a solution of low concentration (or pure solvent) to a solution of high concentration.

Teacher's Note:
1. In osmosis the solvent always moves towards the more concentrated solution.