4: Vapour Pressure

Correct answer: C. 0.526 mol kg⁻¹

Explanation: Take 1 L of solution. Its mass is 1.04 × 1000 = 1040 g. The solution contains 0.5 mol glucose, whose mass is 0.5 × 180 = 90 g. Thus solvent mass = 1040 − 90 = 950 g = 0.950 kg. Molality = 0.5/0.950 = 0.526 mol kg⁻¹ approximately. Option B confuses molarity with molality, while option C correctly uses solvent mass.

Correct answer: A. 18 g mol⁻¹

Explanation: The moles of water are n₁ = 9/18 = 0.50 mol. For the solute, x₂ = n₂/(n₁ + n₂) = 0.02. Solving gives 0.02(0.50 + n₂) = n₂, so n₂ = 0.010/0.98 ≈ 0.0102 mol. Therefore, molar mass = 0.18/0.0102 ≈ 17.6 g mol⁻¹, which is approximately 18 g mol⁻¹. Option A is correct.

Correct answer: B. 19 g mol⁻¹

Explanation: For a dilute solution of a non-volatile solute, relative lowering of vapour pressure equals the solute mole fraction: x₂ = 0.05. The 18 g of water represents 18/18 = 1.00 mol. Let n₂ be solute moles. Then 0.05 = n₂/(1 + n₂), giving n₂ = 0.05/0.95 = 0.0526 mol. Hence, molar mass = 1/0.0526 ≈ 19 g mol⁻¹, so option B is correct.

Correct answer: A. 43.2 g mol−1

Explanation: Assume the solution contains a nonvolatile solute and use the mole-fraction definition xsolute = nsolute/(nsolute + nwater). The 18 g of water corresponds to 18/18 = 1.00 mol water. Therefore 0.04 = n/(1 + n), giving 0.04 + 0.04n = n and n = 0.04/0.96 = 0.04167 mol solute. Its molar mass is mass/moles = 1.8/0.04167 ≈ 43.2 g mol−1.

Correct answer: A. The vapour pressure will not be contributed only by the solvent

Explanation: The usual relative-lowering-of-vapour-pressure method assumes a non-volatile solute, so the vapour pressure of the solution is attributed to the solvent alone. If the solute is volatile, it also enters the vapour phase and contributes its own partial pressure. The simple Raoult’s-law relation is then not applicable in the assumed form, causing an error in the molar-mass calculation unless both components’ vapour pressures are considered.