Question

Vapour pressure of $C_6{H}_6$ and $C_7{H}_8$ mixture at ${50}^{\circ }C$ is given by $P\left(mmHg\right)=179X_B+92$, where $X_B$ is the mole fraction of $C_6{H}_6$. A solution is prepared by mixing $936g$ benzene and $736g$ toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of ${50}^{\circ }C$, what would be the mole fraction of $C_6{H}_6$ in the vapour state? (nearest integer)

Answer 1

$P=179X_B+92$
$P_B^{\circ }=271,P_T^{\circ }=92$
$n_B=\frac{936}{78}=12,n_T=\frac{736}{92}=8$
$X_B=\frac{12}{20}=0.6X_T=0.4$
$P_T=271\times 0.6+92\times 0.4=199.4$
$Y_B=\frac{271\times 0.6}{199.4}=0.815$
$Y_T=0.185$
On further condensation
$X_B=0.185,X_T=0.185$
$P_T=271\times 0.815+92\times 0.185=237.844$
$Y_B=\frac{271\times 0.815}{237.844}=0.9286$