The vapour pressure of pure benzene and methyl benzene at $$27°C$$ is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapour phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is ______ $$\times 10^{-2}$$ (nearest integer)

Correct Answer: 23

Equimolar mixture: $$x_1 = x_2 = 0.5$$.

By Raoult's law: $$P_{benzene} = 0.5 \times 80 = 40$$ Torr. $$P_{toluene} = 0.5 \times 24 = 12$$ Torr.

Total pressure = 52 Torr.

Mole fraction of toluene in vapour = $$12/52 = 3/13 = 0.2307... \approx 23 \times 10^{-2}$$.

The answer is $$\boxed{23}$$.

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources