An aqueous solution containing 1.00 mol kg⁻¹ of sucrose is prepared at 25°C. The activity coefficient of sucrose is $gamma=0.95$ and the vapour pressure of pure water at this temperature is 23.8 mmHg. Calculate the vapour pressure of the solution and the lowering in vapour pressure $DeltaP$.

Step-by-step solution

1. Determine the number of moles of sucrose per kg of water: $n_{\text{sucrose}}=1.00\text{mol}$.
2. Number of moles of water per kg: $n_{\text{water}}=55.5\text{mol}$ (since 1 kg water ≈ 55.5 mol).
3. Total moles in solution: $n_{\text{tot}}=55.5+1.00=56.5\text{mol}$.
4. Mole fraction of water: $x_{\text{w}}=\frac{55.5}{56.5}=0.981$.
5. Effective mole fraction accounting for non‑ideality: $x_{\text{w}}^{\prime }=x_{\text{w}}\gamma =0.981\times 0.95=0.932$.
6. Apply Raoult’s law: $P_{\text{sol}}=x_{\text{w}}^{\prime }{P}^0=0.932\times 23.8\text{mmHg}=22.2\text{mmHg}$.
7. Compute vapour pressure lowering: $\Delta P=P^0-P_{\text{sol}}=23.8-22.2=1.6\text{mmHg}$.

Concept tested: Raoult's law with activity coefficient

Published 5 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked