Q.At $293\ \text{K}$, the vapour pressure of pure benzene is $74.66\ \text{mm Hg}$ and that of pure toluene is $22.31\ \text{mm Hg}$. A solution is prepared by mixing the two liquids such that the mole fraction of toluene is $0.25$. Assuming the solution is ideal, calculate

[!TLDR] Apply Raoult's law to each component, then sum for the total; find the vapour mole fraction from Dalton's law. [!ANSWER] (a) Total vapour pressure $\approx 61.57\ …

(a) By Raoult's law, $p_{benzene} = x_{benzene}\, p^{\circ}_{benzene} = 0.75 \times 74.66 = 55.995\ \text{mm Hg}$ and $p_{toluene} = x_{toluene}\, p^{\circ}_{toluene} = 0.25 \times 22.31 = 5.5775\ \text{mm Hg}$. The total vapour pressure is $p_{total} = 55.995 + 5.5775 = 61.5725 \approx 61.57\ \text{mm Hg}$. (b) By Dalton's law, the mole fraction of benzene in the vapour is $y_{benzene} = p_{benzene}/p_{total} = 55.995/61.5725 \approx 0.909$. [!ANSWER …

Compute each partial pressure via Raoult's law, sum for the total, then divide a component's partial pressur …

Do not use the liquid-phase mole fraction ($x_{benzene}=0.75$) as the vapour-phase composition — the vapour is always enriched in the more volatile componen …