JEE Chemistry Solutions MCQs Set 11

Answer: (d) 6.625 g

Explanation:
1. Moles = molarity × volume (L) = 0.25 × 0.250 = 0.0625 mol.
2. Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g mol⁻¹.
3. Mass = 0.0625 × 106 = 6.625 g.

Teacher's Note:
1. Convert mL to L before multiplying by molarity.
2. Mass = moles × molar mass.

Answer: (d) 0.1

Explanation:
1. Moles of water = 36/18 = 2.
2. Moles of C₂H₅OH = 828/46 = 18.
3. Mole fraction of water = 2/(2 + 18) = 0.1.

Teacher's Note:
1. Mole fraction = moles of the component / total moles.
2. Molar masses: water 18, ethanol 46.

Answer: (b) 3.2

Explanation:
1. Take 1 mole of solution: 0.2 mol solute and 0.8 mol benzene.
2. Mass of benzene = 0.8 × 78 = 62.4 g = 0.0624 kg.
3. Molality = 0.2/0.0624 = 3.2 mol kg⁻¹.

Teacher's Note:
1. Molality uses the mass of solvent in kg.
2. Always pick a convenient basis such as 1 mole of solution.

Answer: (c) 43.2 mm Hg

Explanation:
1. x(heptane) = 1/5 = 0.2 and x(octane) = 4/5 = 0.8.
2. P = 0.2 × 92 + 0.8 × 31 = 18.4 + 24.8 = 43.2 mm Hg.

Teacher's Note:
1. For an ideal solution the total pressure is the sum of the partial pressures given by Raoult's law.
2. Both components contribute because both are volatile.

Answer: (c) 0.27

Explanation:
1. Partial pressures: benzene 0.5 × 160 = 80 torr, toluene 0.5 × 60 = 30 torr.
2. Total = 110 torr.
3. Mole fraction of toluene in vapour = 30/110 = 0.27.

Teacher's Note:
1. The vapour is richer in the more volatile component (benzene).
2. Dalton's law: y = partial pressure / total pressure.

Answer: (c) Negative deviation from Raoult's law

Explanation:
1. A maximum boiling azeotrope boils at a temperature higher than either component.
2. This happens when A–B attractions are stronger than A–A and B–B, so the vapour pressure is lower than the Raoult value: negative deviation.

Teacher's Note:
1. Negative deviation gives maximum boiling azeotropes (ΔH_mix < 0), for example HNO₃ + water.
2. Positive deviation gives minimum boiling azeotropes.

Answer: (c) Shows a positive deviation from Raoult's law

Explanation:
1. A minimum boiling azeotrope has a higher vapour pressure than predicted by Raoult's law.
2. This is positive deviation, which arises when A–B interactions are weaker than A–A and B–B.

Teacher's Note:
1. Example of a minimum boiling azeotrope: ethanol + water.
2. Ideal solutions do not form azeotropes.

Answer: (a) Al₂(SO₄)₃

Explanation:
1. K₄[Fe(CN)₆] gives 4K⁺ + [Fe(CN)₆]⁴⁻, so i = 5.
2. Al₂(SO₄)₃ gives 2Al³⁺ + 3SO₄²⁻, so i = 5.
3. NaCl: 2, Na₂SO₄: 3, Al(NO₃)₃: 4.

Teacher's Note:
1. Count the total number of ions per formula unit.
2. Complex ions stay intact in solution.

Answer: (c) i = 1 − x + x/n

Explanation:
1. Start: 1 mol A. After association x mol A has reacted, so moles of A left = 1 − x.
2. Aₙ formed = x/n mol.
3. Total moles = 1 − x + x/n, and initially 1, so i = 1 − x + x/n.

Teacher's Note:
1. Association lowers i below 1; dissociation raises it above 1.
2. i = moles after / moles before.

Answer: (b) 0.125 mm Hg

Explanation:
1. Moles of sucrose = 100/342 = 0.2924; moles of water = 1000/18 = 55.56.
2. x(sucrose) = 0.2924/(0.2924 + 55.56) = 0.005236.
3. ΔP = P°x = 23.8 × 0.005236 = 0.125 mm Hg.

Teacher's Note:
1. The lowering of vapour pressure ΔP = P° × mole fraction of the solute.
2. Do not forget to convert both masses into moles.

Answer: (b) 206.88

Explanation:
1. ΔP/P° = 0.30/17.54 = 0.01710 = x(solute).
2. n(water) = 100/18 = 5.556 mol, and x = n/(n + 5.556) gives n = 0.01710 × 5.556/(1 − 0.01710) = 0.0967 mol.
3. M = 20/0.0967 = 206.9 g mol⁻¹.

Teacher's Note:
1. Use the exact form ΔP/P° = n/(n + N) when the solution is not very dilute.
2. Molar mass = mass/moles.

Answer: (a) 0.513 °C

Explanation:
1. Latent heat per gram = 9700/18 = 538.9 cal g⁻¹.
2. \( K_b=\frac{RT_b^2}{1000\,L_v}=\frac{1.987\times(373)^2}{1000\times538.9}=0.513\ \text{K kg mol}^{-1} \).

Teacher's Note:
1. Kb = R Tb² M1/(1000 ΔHvap), or with L per gram, R Tb²/(1000 L).
2. Higher latent heat gives a smaller Kb.

Answer: (d) 100

Explanation:
1. ΔTb = Kb m gives m = 0.216/2.16 = 0.1 mol kg⁻¹.
2. m = (0.15/M)/0.015 kg, so 0.15/(0.015 M) = 0.1.
3. M = 0.15/(0.015 × 0.1) = 100 g mol⁻¹.

Teacher's Note:
1. Molality = moles of solute per kg of solvent.
2. Convert 15 g to 0.015 kg.

Answer: (b) −3.53 °C

Explanation:
1. Moles of HBr = 8.1/81 = 0.1; molality = 0.1/0.1 kg = 1 m.
2. HBr ⇌ H⁺ + Br⁻ with α = 0.9 gives i = 1 + α = 1.9.
3. ΔTf = i Kf m = 1.9 × 1.86 × 1 = 3.53 K, so Tf = −3.53 °C.

Teacher's Note:
1. i = 1 + (n − 1)α for a compound giving n ions.
2. Freezing point = 0 − ΔTf for aqueous solutions.

Answer: (a) 256 g mol⁻¹

Explanation:
1. \( M=\frac{1000\,K_f\,w}{\Delta T_f\,W}=\frac{1000\times5.12\times1.00}{0.40\times50}=256\ \text{g mol}^{-1} \).

Teacher's Note:
1. W is the mass of solvent in grams in this form of the formula.
2. Check the order of magnitude of the answer.

Answer: (a) −0.45 °C

Explanation:
1. HX ⇌ H⁺ + X⁻ with α = 0.2 gives i = 1.2.
2. ΔTf = 1.2 × 1.86 × 0.2 = 0.4464 K.
3. Freezing point = −0.45 °C.

Teacher's Note:
1. Weak acid: i = 1 + α.
2. Remember the sign: the freezing point is lowered.

Answer: (c) 68.4

Explanation:
1. Isotonic solutions have equal molar concentration (non-electrolytes).
2. 5 g/342 per 100 mL equals 1 g/M per 100 mL, so M = 342/5 = 68.4.

Teacher's Note:
1. Isotonic means equal osmotic pressure, so equal molarity for non-electrolytes.
2. Both solutions are % w/v, so the same volume is implied.

Answer: (c) 4.8 atm

Explanation:
1. π = CRT = 0.2 × 0.0821 × 293 = 4.81 atm.

Teacher's Note:
1. Use π = iCRT, with i = 1 for non-electrolytes.
2. Temperature must be in kelvin.

Answer: (a) 348.9

Explanation:
1. Molarity of urea = 8.6/60 = 0.1433 mol L⁻¹.
2. 5% solution = 50 g per litre; equal molarity gives 50/M = 0.1433.
3. M = 50/0.1433 = 348.9.

Teacher's Note:
1. A 5% w/v solution has 5 g per 100 mL, that is 50 g per litre.
2. Isotonic means the same concentration in mol L⁻¹.

Answer: (c) \( 0.33\times10^{-2} \)

Explanation:
1. π = CRT, so C = π/(RT) = 0.0821/(0.0821 × 300) = 1/300.
2. C = 3.33 × 10⁻³ = 0.33 × 10⁻² mol L⁻¹.

Teacher's Note:
1. Check the power of ten carefully: 1/300 is about 0.0033.
2. Use π = CRT for a non-electrolyte.