CBSE Class 12 Chemistry Solutions MCQs Set 17

Answer: (d) constant boiling mixtures

Explanation:
1. Azeotropes have the same composition in liquid and vapour phase and boil at a constant temperature.

Teacher's Note:
1. Because vapour and liquid have the same composition, fractional distillation cannot separate an azeotrope.

Answer: (b) 0.186

Explanation:
1. ΔT_f = K_f × m = 1.86 × 0.1 = 0.186

Teacher's Note:
1. For a non-electrolyte i = 1; for an electrolyte multiply by i.

Answer: (c) decrease the freezing point of water

Explanation:
1. A non-volatile solute lowers the vapour pressure of water, which raises its boiling point and lowers its freezing point.
2. Only option (c) is correct.

Teacher's Note:
1. Lower vapour pressure leads to both effects: higher boiling point and lower freezing point.

Answer: (c) a hypertonic solution

Explanation:
1. In a hypertonic solution water moves out of the cell by osmosis, so the cell shrinks (plasmolysis).

Teacher's Note:
1. Water always moves from lower solute concentration to higher solute concentration across the membrane.

Answer: (d) below 0°C

Explanation:
1. A dissolved solute depresses the freezing point of water, so the solution freezes below 0°C.

Teacher's Note:
1. Any solute, electrolyte or not, lowers the freezing point; it never raises it.

Answer: (b) 87%

Explanation:
1. Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻; n = 3
2. α = \( \frac{i-1}{n-1} \) = \( \frac{2.74-1}{3-1} \) = 0.87 = 87%

Teacher's Note:
1. Count the ions produced per formula unit carefully; here n = 3.

Answer: (c) \( \frac{M}{10} \) BaCl₂

Explanation:
1. π = iCRT; at the same concentration, the solute with the largest i gives the highest π.
2. BaCl₂ gives 3 ions (i = 3), NaCl gives 2, urea and glucose 1.

Teacher's Note:
1. Osmotic pressure is proportional to the total concentration of particles, not just the molarity of the solute.

Answer: (c) 0.5 M Urea solution

Explanation:
1. Isotonic solutions have the same particle concentration. Glucose and urea are non-electrolytes (i = 1), so 0.5 M urea matches 0.5 M glucose.

Teacher's Note:
1. For isotonic solutions compare i × C, not just C.

Answer: (b) It increases with temperature

Explanation:
1. The value of K_H increases with rise in temperature, so the solubility of a gas decreases on heating.

Teacher's Note:
1. Higher K_H means lower solubility – that is why aquatic species are more comfortable in cold water.

Answer: (a) osmotic pressure

Explanation:
1. Osmotic pressure is measured at room temperature and gives appreciable values even for very dilute solutions, so it suits polymers.

Teacher's Note:
1. Polymers have huge molar masses, so ΔT_b and ΔT_f would be too small to measure; π is still measurable.

Answer: (b) molality

Explanation:
1. Molality uses mass of solvent, which does not change with temperature; the others use volume, which changes.

Teacher's Note:
1. Any concentration unit that has volume in it changes with temperature.

Answer: (d) 528 g

Explanation:
1. C = K × p = 3 × 10⁻² × 4 = 0.12 mol L⁻¹
2. Moles in 100 L = 12 mol; mass = 12 × 44 = 528 g

Teacher's Note:
1. Check the units of K: here it is given as mol L⁻¹ atm⁻¹, so solubility = K × p directly.

Answer: (b) exothermic reaction

Explanation:
1. Stronger attractions form on mixing (negative deviation), so heat is released; the mixing is exothermic.

Teacher's Note:
1. Negative deviation goes with ΔH_mix < 0 and ΔV_mix < 0.

Answer: (d) C₇H₁₆–C₈H₁₈

Explanation:
1. n-Heptane and n-octane are similar non-polar molecules with nearly equal intermolecular forces, so they form a nearly ideal solution.

Teacher's Note:
1. Ideal behaviour is expected when the two components are chemically alike in structure and polarity.

Answer: (c) 3

Explanation:
1. K₂[HgI₄] → 2K⁺ + [HgI₄]²⁻; 3 ions, so i = 3.

Teacher's Note:
1. The complex ion stays as one unit; do not split it into Hg²⁺ and I⁻.

Answer: (c) 244

Explanation:
1. Benzoic acid (M = 122) dimerises in benzene by hydrogen bonding, so its observed molar mass is 2 × 122 = 244.

Teacher's Note:
1. Association gives i < 1 and an abnormally high molar mass.

Answer: (a) 18% (W/V) solution of glucose

Explanation:
1. 6% urea = 60 g L⁻¹ = 60/60 = 1 M.
2. 18% glucose = 180 g L⁻¹ = 180/180 = 1 M; both are non-electrolytes, so they are isotonic.

Teacher's Note:
1. Convert % (W/V) to molarity first (g per litre ÷ molar mass), then compare particle concentrations.

Answer: (b) acetone + C₂H₅OH

Explanation:
1. In ethanol + acetone, acetone breaks some hydrogen bonds between ethanol molecules, so A–B attractions are weaker and the solution shows positive deviation.
2. The other three mixtures form stronger attractions on mixing and show negative deviation.

Teacher's Note:
1. Mixtures that break hydrogen bonds on mixing show positive deviation; mixtures that form new hydrogen bonds show negative deviation.

Answer: (d) A–S, B–R, C–Q, D–P

Explanation:
1. i = 1: no dissociation or association; i > 1: dissociation; i < 1: association; i = 0 would mean no particles at all, which is impossible.

Teacher's Note:
1. Remember: dissociation increases the number of particles (i > 1), association decreases it (i < 1).

Answer: (b) 0.459

Explanation:
1. In 100 g solution: benzene = 30 g, CCl₄ = 70 g.
2. n(C₆H₆) = \( \frac{30}{78} \) = 0.385 mol; n(CCl₄) = \( \frac{70}{154} \) = 0.455 mol
3. x(C₆H₆) = \( \frac{0.385}{0.385 + 0.455} \) = 0.459

Teacher's Note:
1. Mass % always means grams per 100 g of solution, so the solvent mass is 100 − 30 = 70 g.

Answer: (d) 17.95 m

Explanation:
1. Molar mass of ethylene glycol, C₂H₆O₂ = 62 g mol⁻¹; moles = \( \frac{222.6}{62} \) = 3.59 mol
2. Molality = \( \frac{3.59}{0.200\ \text{kg}} \) = 17.95 m

Teacher's Note:
1. Molality uses the mass of the solvent in kg, not the mass of the solution.

Answer: (a) 4.575 g

Explanation:
1. Moles = M × V(L) = 0.15 × 0.250 = 0.0375 mol
2. Molar mass of C₆H₅COOH = 122 g mol⁻¹; mass = 0.0375 × 122 = 4.575 g

Teacher's Note:
1. Convert mL to L before multiplying by molarity.

Answer: (b) 55.5 M

Explanation:
1. Molarity is number of moles per litre.
2. 900 g H₂O = \( \frac{900}{18} \) moles = 50 moles; 900 g H₂O = 900 mL of H₂O
3. ∴ Molarity = \( \frac{50}{900} \) × 1000 = 55.5 M

Teacher's Note:
1. The molarity of pure water is always about 55.5 M, whatever the amount taken, since 1 L of water has 1000/18 mol.

Answer: (a) (i) and (iii) at constant T

Explanation:
1. At constant temperature, the solubility of a gas depends on the nature of the gas (its K_H) and on its partial pressure (Henry's law).

Teacher's Note:
1. Read the condition 'at constant T' carefully – it is what makes option (a) the complete answer.

Answer: (a) mole fraction

Explanation:
1. Raoult's law relates vapour pressure to mole fraction: p_A = p_A° x_A.

Teacher's Note:
1. Vapour pressure laws (Raoult's and Henry's) are always written in mole fractions.

Answer: (c) 17.44 mm Hg

Explanation:
1. n(glucose) = \( \frac{25}{180} \) = 0.139 mol; n(water) = \( \frac{450}{18} \) = 25 mol
2. x(water) = \( \frac{25}{25 + 0.139} \) = 0.9945
3. p = p° x(water) = 17.535 × 0.9945 = 17.44 mm Hg

Teacher's Note:
1. A non-volatile solute only slightly lowers vapour pressure; the answer must be a little less than 17.535 mm Hg.

Answer: (d) 0.2

Explanation:
1. In solution, if x_A = x, x_B = 2x and if p_A° = p, p_B° = 2p
2. ∴ p_A = x × p, p_B = 2x × 2p = 4x × p
3. ∴ P_total = 5xp
4. Mole fraction in vapour phase (y_A) = \( \frac{p_A}{P_{total}} \) = \( \frac{xp}{5xp} \) = \( \frac{1}{5} \) = 0.2

Teacher's Note:
1. Mole fraction in vapour = partial pressure / total pressure (Dalton's law); the more volatile component is richer in the vapour.

Answer: (d) 0 ≤ x₂ ≤ 1

Explanation:
1. When one component follows Raoult's law, the solution is ideal. The second component will also follow Raoult's law at all concentrations.

Teacher's Note:
1. An ideal solution is ideal as a whole – both components obey Raoult's law over the full composition range.

Answer: (c) does not obey Raoult's law over entire range of concentration

Explanation:
1. An ideal solution obeys Raoult's law over the entire range of concentration, with ΔH_mix = 0 and ΔV_mix = 0.
2. So statement (c) is incorrect.

Teacher's Note:
1. Watch for 'incorrect' in the stem – pick the false statement.

Answer: (b) shows negative deviation

Explanation:
1. According to Raoult's law,
2. p_A = x_A × p_A° = \( \frac{1}{3} \) × 45 torr = 15 torr
3. p_B = x_B × p_B° = \( \frac{2}{3} \) × 36 torr = 24 torr
4. ∴ Pressure expected by Raoult's law = 15 + 24 = 39 torr
5. Thus, observed pressure (38 torr) is less than expected value. Hence, the solution shows negative deviation.

Teacher's Note:
1. Compare the observed total pressure with the Raoult's-law value: lower means negative deviation, higher means positive.

Answer: (b) Number of particles of the solute in solution

Explanation:
1. Colligative properties depend only on the number of solute particles in a given amount of solvent, not on their nature.

Teacher's Note:
1. 'Colligative' means 'bound together' – think 'count of particles', not identity.

Answer: (d) 64

Explanation:
1. \( \frac{p^\circ - p_s}{p^\circ} \) = \( \frac{n_2}{n_1} \) = \( \frac{w_2/M_2}{w_1/M_1} \)
2. \( \frac{185 - 183}{185} \) = \( \frac{1.2/M_2}{100/58} \) (Molar mass of CH₃COCH₃ = 58 g mol⁻¹)
3. = \( \frac{1.2 \times 58}{100 \times M_2} \) or M₂ = \( \frac{1.2 \times 58}{100} \) × \( \frac{185}{2} \) = 64.38 g mol⁻¹ ≈ 64

Teacher's Note:
1. For dilute solutions, n₁ + n₂ ≈ n₁, which is why the simpler ratio w₂M₁/w₁M₂ can be used.

Answer: (b) 3 g

Explanation:
1. \( \frac{p^\circ - p_s}{p^\circ} \) = \( \frac{n_2}{n_1} \) = \( \frac{w_2 M_1}{w_1 M_2} \)
2. As (p° − p_s)/p° is same in the two cases, \( \left(\frac{w_2 M_1}{w_1 M_2}\right)_{glucose} = \left(\frac{w_2 M_1}{w_1 M_2}\right)_{urea} \)
3. \( \frac{w_2 \times 18}{50 \times 180} \) = \( \frac{1 \times 18}{50 \times 60} \) or w₂ = 3 g

Teacher's Note:
1. Same solvent and same amount: equal lowering simply needs equal moles of solute (1/60 mol here).

Answer: (a) Cellulose acetate

Explanation:
1. Films of cellulose acetate, permeable to water but impermeable to impurities and ions, are used as the membrane in reverse osmosis (desalination) plants.

Teacher's Note:
1. Remember cellulose acetate as the standard RO membrane in desalination questions.

Answer: (d) it loses water due to osmosis

Explanation:
1. The salt solution outside is more concentrated than the cell sap, so water moves out of the mango by osmosis and it shrivels.

Teacher's Note:
1. No external pressure is applied, so this is ordinary osmosis (exosmosis), not reverse osmosis.

Answer: (b) 18°C

Explanation:
1. π = CRT. Thus, a plot of π vs C will be linear with slope = RT. Hence,
2. RT = 291 R or T = 291 K = (291 − 273)°C = 18°C

Teacher's Note:
1. Check the units of the options: 291 K and 18°C are the same temperature.

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

Explanation:
1. Both lowering of vapour pressure and osmotic pressure are proportional to the concentration of solute particles, so they are proportional to each other; A is true.
2. R is also true, but being a colligative property alone does not explain the proportionality stated in A.

Teacher's Note:
1. All colligative properties depend on the number of solute particles, which links them to one another.

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

Explanation:
1. Polymers have very high molar masses, so ΔT_b and ΔT_f of their solutions are too small to be measured accurately; A is true.
2. R is false; the reason is the very small magnitude of these changes. Osmotic pressure is used instead.

Teacher's Note:
1. Large molar mass means few moles of solute for a given mass, hence tiny elevation/depression values.

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

Explanation:
1. KCl gives two ions (i = 2) while urea does not dissociate (i = 1). Since elevation of boiling point depends on the number of particles, KCl solution boils at a higher temperature; R explains A.

Teacher's Note:
1. Always apply the van’t Hoff factor before comparing colligative properties of solutions of equal concentration.

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

Explanation:
1. Azeotropes arise only from large deviations from Raoult’s law, so only non-ideal solutions form them.
2. They cannot be separated by fractional distillation because liquid and vapour have the same composition; this is true but does not explain why only non-ideal solutions form them.

Teacher's Note:
1. Minimum boiling azeotropes come from positive deviation and maximum boiling azeotropes from negative deviation.

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

Explanation:
1. K_H = p/x, so the assertion is true.
2. Correct R. Henry's law constant depends upon temperature.

Teacher's Note:
1. K_H increases with temperature, which is why gases are less soluble in warm water.

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

Explanation:
1. Correct A. A mixture of chloroform and acetone forms a non-ideal solution with negative deviation.
2. Acetone (b.p. 329 K) is more volatile than chloroform (b.p. 334 K), so R is true.

Teacher's Note:
1. Chloroform and acetone form a hydrogen bond (Cl₃C–H···O=C), the classic example of negative deviation.

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

Explanation:
1. Correct R. On applying higher pressure on the solution, the solvent will flow from solution into the pure solvent through the semipermeable membrane.

Teacher's Note:
1. Only the solvent ever passes through a semipermeable membrane – in osmosis and in reverse osmosis.

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

Explanation:
1. Correct A. 0.1 M HCl and 0.1 M NaCl will have the same osmotic pressure because HCl and NaCl dissociate to produce two particles each.
2. H⁺ ions do have a greater ionic mobility than Na⁺ ions, so R is true (but mobility does not affect osmotic pressure).

Teacher's Note:
1. Colligative properties count particles; ionic mobility affects conductance, not osmotic pressure.

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

Explanation:
1. Correct explanation. Due to high molar masses of polymers, the elevation in boiling point or depression in freezing point are so low that they cannot be measured accurately.

Teacher's Note:
1. The real reason is the size of the effect: only osmotic pressure is large enough to measure for macromolecules.

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

Explanation:
1. Correct R. It depends upon the nature of the solvent because K_f or K_b values depend upon the solvent.

Teacher's Note:
1. ΔT_b depends on the number of solute particles and on the solvent (through K_b), not on the identity of the solute.