RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
A solution is more than salt stirred into water — its freezing point, boiling point and vapour pressure all shift in ways you can predict with a formula. This chapter shows you how Raoult's law governs vapour pressure, how the four colligative properties let you weigh a molecule you can never see, and why the van't Hoff factor tells you whether a solute has split apart or clumped together.
Equimolal aqueous solutions of glucose, KCl, K₂SO₄ and AlCl₃ are cooled. Which one freezes at the lowest temperature?
Answer
AlCl₃. Depression of freezing point is ΔTf = i Kf m, so the lowest freezing point belongs to the solute with the largest van't Hoff factor. Glucose gives i ≈ 1, KCl gives i ≈ 2, K₂SO₄ gives i ≈ 3 and AlCl₃ dissociates into Al³⁺ and 3Cl⁻, giving i ≈ 4. With m and Kf identical, AlCl₃ produces the greatest ΔTf and therefore the lowest freezing temperature.
A mixture of ethanol and acetone shows a vapour pressure higher than that predicted by Raoult's law. This solution will:
Answer
Form a minimum-boiling azeotrope and have ΔH_mix positive. Higher-than-expected vapour pressure is a positive deviation from Raoult's law: the new ethanol–acetone interactions are weaker than the hydrogen bonding within pure ethanol, so molecules escape more easily. Breaking those stronger original interactions absorbs heat (ΔH_mix positive), volume increases slightly, and the composition of maximum vapour pressure boils at the lowest temperature — a minimum-boiling azeotrope.
The Henry's law constant for a gas in water is 4.8 ×10⁴ bar at 298 K and 7.6 ×10⁴ bar at 318 K. Which conclusion is correct?
Answer
The gas is less soluble at 318 K because solubility varies inversely with K_H. From p = K_H x, the mole fraction dissolved at a fixed partial pressure is x = p/K_H, so a larger K_H means a smaller x. Dissolution of a gas is exothermic, so raising the temperature drives gas out of solution and pushes K_H up.
0.90 g of a non-volatile solute dissolved in 100 g of water lowers the freezing point by 0.093 °C (Kf = 1.86 K kg mol⁻¹). The molar mass of the solute is:
Answer
180 g mol⁻¹. Using M_B = (1000 × Kf × w_B)/(ΔTf × w_A) = (1000 × 1.86 × 0.90)/(0.093 × 100) = 1674/9.3 = 180 g mol⁻¹. Check: molality = ΔTf/Kf = 0.093/1.86 = 0.05 mol kg⁻¹, so moles in 0.100 kg water = 0.005, and 0.90/0.005 = 180 g mol⁻¹.
Assertion (A): Osmotic pressure is preferred over depression of freezing point for finding the molar mass of a protein. Reason (R): Osmotic pressure has a large measurable magnitude even in very dilute solution and is measured at room temperature, where a protein does not decompose.
Answer
Both A and R are true and R is the correct explanation of A. A protein of high molar mass gives an extremely small number of moles per kg of solvent, so ΔTf would be far too tiny to measure accurately. Osmotic pressure, π = CRT, produces a readable pressure even for a 10⁻³ M solution, and because the measurement is made at ordinary temperature the delicate protein is not denatured — exactly the reason stated in R.
At 300 K the vapour pressures of pure liquids A and B are 120 mm Hg and 40 mm Hg. A solution is made by mixing 2 mol of A with 3 mol of B. Calculate the total vapour pressure and the mole fraction of A in the vapour phase.
Answer
Mole fractions in the liquid: x_A = 2/(2+3) = 0.4 and x_B = 0.6. Partial pressures by Raoult's law: p_A = p°_A x_A = 120 × 0.4 = 48 mm Hg; p_B = p°_B x_B = 40 × 0.6 = 24 mm Hg. Total vapour pressure p_total = 48 + 24 = 72 mm Hg. Mole fraction of A in the vapour (Dalton's law) y_A = p_A/p_total = 48/72 = 0.67. The vapour is richer in A (0.67) than the liquid (0.40) because A is the more volatile component.
0.585 g of sodium chloride (molar mass 58.5 g mol⁻¹) is dissolved in 100 g of water. The observed depression of freezing point is 0.353 K. Calculate the van't Hoff factor and the degree of dissociation of NaCl. (Kf = 1.86 K kg mol⁻¹)
Answer
Step 1 — molality: moles of NaCl = 0.585/58.5 = 0.01 mol in 0.100 kg water, so m = 0.10 mol kg⁻¹. Step 2 — expected depression if NaCl did not dissociate: ΔTf(calc) = Kf × m = 1.86 × 0.10 = 0.186 K. Step 3 — van't Hoff factor: i = ΔTf(obs)/ΔTf(calc) = 0.353/0.186 = 1.90. Step 4 — degree of dissociation: NaCl → Na⁺ + Cl⁻ gives n = 2, so α = (i − 1)/(n − 1) = (1.90 − 1)/(2 − 1) = 0.90. NaCl is about 90% dissociated at this concentration; i is slightly below the ideal value of 2 because of interionic attraction in solution.
Define an ideal solution and state two conditions it must satisfy. Give one example.
Answer
An ideal solution is one in which every component obeys Raoult's law over the whole range of composition, so p_A = p°_A x_A for each component. The two conditions are: (i) the enthalpy of mixing is zero, ΔH_mix = 0, since no heat is absorbed or released; and (ii) the volume of mixing is zero, ΔV_mix = 0, so the total volume equals the sum of the component volumes. Both follow from the solute–solvent interaction being identical in strength to the solute–solute and solvent–solvent interactions. Example: a mixture of n-hexane and n-heptane, whose molecules experience nearly identical dispersion forces.
(i) State Raoult's law for a solution of a non-volatile solute in a volatile solvent and derive the expression for relative lowering of vapour pressure. (ii) 4.50 g of a non-volatile, non-electrolyte solute is dissolved in 150 g of benzene. The solution boils 0.253 °C above pure benzene. Calculate the molar mass of the solute. (Kb for benzene = 2.53 K kg mol⁻¹)
Answer
(i) For a non-volatile solute, only the solvent contributes to the vapour pressure, so Raoult's law reads p_A = p°_A x_A, where p°_A is the vapour pressure of the pure solvent and x_A its mole fraction in solution. Since x_A + x_B = 1, p_A = p°_A(1 − x_B). Rearranging: p°_A − p_A = p°_A x_B, and dividing by p°_A gives (p°_A − p_A)/p°_A = x_B. The left side is the relative lowering of vapour pressure; it equals the mole fraction of solute and is therefore a colligative property, depending on the number of solute particles and not on their nature. For a dilute solution x_B ≈ n_B/n_A = (w_B/M_B)(M_A/w_A), which gives M_B = (w_B M_A p°_A)/(w_A Δp). (ii) Step 1 — molality from ΔTb = Kb m: m = ΔTb/Kb = 0.253/2.53 = 0.100 mol kg⁻¹. Step 2 — moles of solute in 150 g = 0.150 kg of benzene: n_B = 0.100 × 0.150 = 0.0150 mol. Step 3 — molar mass: M_B = w_B/n_B = 4.50/0.0150 = 300 g mol⁻¹. Direct formula check: M_B = (1000 × Kb × w_B)/(ΔTb × w_A) = (1000 × 2.53 × 4.50)/(0.253 × 150) = 11385/37.95 = 300 g mol⁻¹.
(i) Explain the terms osmosis, osmotic pressure, isotonic solution and reverse osmosis. (ii) 1.20 g of a polymer is dissolved in water to make 250 mL of solution. Its osmotic pressure at 300 K is found to be 2.46 ×10⁻³ bar. Calculate the molar mass of the polymer. (R = 0.083 L bar K⁻¹ mol⁻¹)
Answer
(i) Osmosis is the net flow of solvent molecules through a semipermeable membrane from a region of lower solute concentration to one of higher solute concentration. Osmotic pressure (π) is the excess pressure that must be applied on the solution side to just stop this flow; for a dilute solution π = CRT. Two solutions with the same osmotic pressure at the same temperature are isotonic — a 0.9% w/V NaCl drip is isotonic with blood cells, so the cells neither swell nor shrink. Reverse osmosis occurs when a pressure greater than π is applied to the solution: solvent is then forced backwards through the membrane, out of the concentrated side, which is how sea water is desalinated. (ii) Step 1 — write π = (n/V)RT = (w_B/M_B)(RT/V), so M_B = w_B RT/(π V). Step 2 — substitute the data: w_B = 1.20 g, R = 0.083 L bar K⁻¹ mol⁻¹, T = 300 K, π = 2.46 ×10⁻³ bar, V = 0.250 L. Numerator: 1.20 × 0.083 × 300 = 29.88 L bar g mol⁻¹. Denominator: 2.46 ×10⁻³ × 0.250 = 6.15 ×10⁻⁴ L bar. Step 3 — M_B = 29.88/(6.15 ×10⁻⁴) = 4.86 ×10⁴ g mol⁻¹. The polymer has a molar mass of about 48 600 g mol⁻¹. Such a large value would give ΔTf of only about 0.00009 K, which is why osmotic pressure was used instead.
A food technologist is testing two anti-freeze mixtures for a cold-storage coolant. Sample X is made by dissolving 12.4 g of ethylene glycol (M = 62 g mol⁻¹) in 200 g of water. Sample Y is made by dissolving 11.7 g of NaCl (M = 58.5 g mol⁻¹) in the same mass of water; NaCl is found to be 85% dissociated. Take Kf(water) = 1.86 K kg mol⁻¹. (a) Calculate the freezing point of sample X. (b) Calculate the van't Hoff factor for NaCl in sample Y. (c) Calculate the freezing point of sample Y. (d) Which sample is the better coolant, and state one reason why ΔTf, and not the mass of solute added, decides the answer.
Answer
(a) Moles of ethylene glycol = 12.4/62 = 0.200 mol in 0.200 kg water, so m = 1.00 mol kg⁻¹. Ethylene glycol is a non-electrolyte, so i = 1 and ΔTf = Kf m = 1.86 × 1.00 = 1.86 K. Freezing point = 0 − 1.86 = −1.86 °C. (b) NaCl → Na⁺ + Cl⁻, so n = 2 and α = 0.85. Using i = 1 + (n − 1)α = 1 + (1)(0.85) = 1.85. (c) Moles of NaCl = 11.7/58.5 = 0.200 mol in 0.200 kg water, so m = 1.00 mol kg⁻¹. ΔTf = i Kf m = 1.85 × 1.86 × 1.00 = 3.44 K. Freezing point = 0 − 3.44 = −3.44 °C. (d) Sample Y is the better coolant because it stays liquid down to −3.44 °C, nearly twice the depression given by X at the same molality. Freezing point depression is a colligative property: it depends on the number of solute particles in solution, not on their identity or mass. NaCl releases about 1.85 particles per formula unit while ethylene glycol releases one, so even at equal molality the ionic solute wins.
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