Question

Benzene and toluene form an ideal solution. Vapour Pressure of pure benzene is 100 torr while that of pure toluene is 50 torr. If mole fraction of benzene in liquid phase is $\frac{1}{3}$, calculate the mole fraction of benzene in vapour phase.

• A. $\frac{2}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{5}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{2}$

According to Raoult's law, the total vapour pressure of a solution is given by
$P_S$ $=$ $P_A^0$${\chi }_A$ $+$ $P_B^0$${\chi }_B$
Given,
Vapour pressure of pure benzene $P_A^0=100$ torr
Vapour pressure of pure toluene $P_B^0=50$ torr
${\chi }_B=\frac{1}{3}$ and ${\chi }_A=\frac{2}{3}$
$P_S$ $=$ $100$$\times$$\frac{1}{3}$ $+$ $50$$\times$$\frac{2}{3}$$=$$\frac{200}{3}$

Mole fraction of Benzene, $Y_B$, in vapour phase is the ratio of partial pressure of Benzene to the total pressure of solution.
$Y_B$ $=$$\frac{\frac{100}{3}}{\frac{200}{3}}$ $=$ $\frac{1}{2}$