Explain why a $1{\text{mol kg}}^{-1}$ aqueous solution of sucrose shows a lower vapour pressure than that predicted by Raoult's law. In your answer discuss the role of the non‑volatile solute, solute–solvent interactions, and the activity coefficient.

Step-by-step solution

1. State Raoult's law for an ideal solution: $p_{\text{solution}}=x_{\text{water}}{p}_{\text{water}}^{\circ }$, where $x_{\text{water}}$ is the mole fraction of water.
2. Calculate the ideal mole fraction for a $1{\text{mol kg}}^{-1}$ sucrose solution (approximately $x_{\text{water}}\approx 0.945$), which would give a predicted vapour pressure $p_{\text{ideal}}=0.945p_{\text{water}}^{\circ }$.
3. Explain that sucrose is non‑volatile, so it does not contribute directly to the vapour pressure; only the reduction in $x_{\text{water}}$ can lower $p$.
4. Introduce the concept of activity: $p_{\text{solution}}={\gamma }_{\text{water}}{x}_{\text{water}}{p}_{\text{water}}^{\circ }$.
5. Describe how strong hydrogen‑bonding between sucrose (with many –OH groups) and water molecules leads to a tighter association than in pure water.
6. Because of these strong solute–solvent interactions, the escaping tendency of water molecules is further reduced, giving ${\gamma }_{\text{water}}<1$ (negative deviation).
7. Consequently, the observed vapour pressure $p_{\text{observed}}={\gamma }_{\text{water}}{x}_{\text{water}}{p}_{\text{water}}^{\circ }$ is lower than $p_{\text{ideal}}$ predicted by Raoult's law.
8. Summarise: the combined effect of (i) dilution of water by a non‑volatile solute and (ii) attractive sucrose–water interactions (activity coefficient < 1) explains the lower vapour pressure.

Concept tested: Raoult's law and deviations

Published 31 August 2026 · How this solution was prepared and checked

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