A 1.20 mol·kg⁻¹ aqueous solution of sucrose at 25°C has an activity coefficient γ = 0.92. The vapor pressure of pure water at this temperature is 23.8 mmHg. Calculate (a) the vapor pressure of the solution and (b) the lowering of vapor pressure ΔP.

hard 4 marks Class 12 · Chemistry · Solutions

Answer

(a) P_solution = 23.34 mmHg; (b) ΔP = 0.46 mmHg (approximately)

The correct answer is (a) P_solution = 23.34 mmHg; (b) ΔP = 0.46 mmHg (approximately) — State the final answers for (a) and (b).

Step-by-step solution

Assume 1 kg of water as the solvent for convenience.
Moles of water in 1 kg: n_water = 1000 g / 18.015 g mol⁻¹ = 55.51 mol.
Moles of sucrose: n_solute = molality × mass of solvent = 1.20 mol·kg⁻¹ × 1 kg = 1.20 mol.
Mole fraction of sucrose: x_solute = n_solute / (n_solute + n_water) = 1.20 / (1.20 + 55.51) = 0.0212.
For a non‑ideal solution, the vapor pressure lowering is given by ΔP = γ_solute × x_solute × P°_water.
Calculate ΔP: ΔP = 0.92 × 0.0212 × 23.8 mmHg = 0.464 mmHg ≈ 0.46 mmHg.
Vapor pressure of the solution: P_solution = P°_water – ΔP = 23.8 mmHg – 0.46 mmHg = 23.34 mmHg.
State the final answers for (a) and (b).

Common mistake: Using molarity instead of molality to obtain mole fraction, leading to an incorrect ΔP.

Concept tested: Vapour pressure lowering with activity coefficient