Question

Calculate the mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50. The vapour pressure of pure benzene is 119 torr; that of toluene is 37 torr at the same temperature. ( If the answer is $x$, then write $x\times {10}^{-3}$)(rounding off to 3 digits)

Answer 1

The partial pressure of toluene in the vapour phase is

$P_T=P_T^0{X}_T=37torr\times 0.5=18.5torr$.

Here, $P_T^0$ is the vapour pressure of pure toluene and $X_T$ is the mole fraction of toluene in the liquid phase.

The partial pressure of benzene in the vapour phase is

$P_B=P_B^0{X}_B=119torr\times 0.5=59.5torr$.

Here, $P_B^0$ is the vapour pressure of pure benzene and $X_B$ is the mole fraction of benzene in the liquid phase.

The total pressure is $P_T+P_B=18.5torr+59.5torr=78torr$.

The mole fraction of toluene in the vapour phase is $Y_T$.

It is given by the expression $Y_T=\frac{P_T}{totalpressure}=\frac{18.5}{78}=0.237$.

Hence, the mole fraction of toluene in the vapour phase is 0.237.

Hence the correct answer is 237