A 0.80\;\text{mol·kg^{-1}} aqueous solution of urea at 25\;^{\circ}\text{C} shows a vapour pressure of 22.5\;\text{mmHg}. The vapour pressure of pure water at this temperature is 23.8\;\text{mmHg}. Assuming urea is a non‑volatile solute, calculate the activity coefficient \gamma_{w} of water in the solution.
Answer
\gamma_{w}=0.96
The correct answer is \gamma_{w}=0.96 — Re‑arrange the vapour‑pressure equation to solve for \gamma_{w}: \gamma_{w}=P/(X_{w}\,P^{\circ})=22.5\;\text{mmHg}/(0.9858\times23.8\;\text{mmHg})=0.959\approx0.96.
Step-by-step solution
- Write the relation including the activity coefficient: P = \gamma_{w}\,X_{w}\,P^{\circ}, where P is the measured vapour pressure, P^{\circ} is the vapour pressure of pure water, and X_{w} is the mole fraction of water.
- Calculate the mole fraction of water. For a 1\;kg solvent, the number of moles of water is n_{w}=1000\;\text{g}/18.015\;\text{g·mol^{-1}}=55.51\;\text{mol}. The solute moles are n_{s}=m=0.80\;\text{mol}. Hence X_{w}=n_{w}/(n_{w}+n_{s})=55.51/(55.51+0.80)=0.9858.
- Re‑arrange the vapour‑pressure equation to solve for \gamma_{w}: \gamma_{w}=P/(X_{w}\,P^{\circ})=22.5\;\text{mmHg}/(0.9858\times23.8\;\text{mmHg})=0.959\approx0.96.
Common mistake: Using P = directly without accounting for the activity coefficient of water
Concept tested: Activity coefficient from vapour‑pressure lowering
Published · Last updated · How this solution was prepared and checked
← All Solutions questions