Vapour pressure of pure benzene is 119 torr and that of toluene is $37.0$ torr at the same temperature. Mole…
Vapour pressure of pure benzene is 119 torr and that of toluene is $37.0$ torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$, will be :
$0.137$
$0.237$
$0.435$
$0.205$
Solution
$\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^0 \times x_{\mathrm{A}}=$ total pressure $\times \mathrm{y}_{\mathrm{A}}$
$
\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^0 \times x_{\mathrm{B}}=\text { total pressure } \times \mathrm{y}_{\mathrm{B}}
$
where $\mathrm{x}$ and $\mathrm{y}$ represents mole fraction in liquid and vapour phase respectively.
$
\frac{\mathrm{P}_{\mathrm{B}}^0 x_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}^0 x_{\mathrm{A}}}=\frac{\mathrm{y}_{\mathrm{B}}}{\mathrm{y}_{\mathrm{A}}} ; \frac{\mathrm{P}_{\mathrm{B}}^0\left(1-x_{\mathrm{A}}\right)}{\mathrm{P}_{\mathrm{A}}^0 x_{\mathrm{A}}}=\frac{1-\mathrm{y}_{\mathrm{A}}}{\mathrm{y}_{\mathrm{A}}}
$
on putting values $\frac{119(1-0.50)}{37 \times 0.50}=\frac{1-y_{\mathrm{A}}}{\mathrm{y}_{\mathrm{A}}}$ on solving $\mathrm{y}_{\mathrm{A}}=0.237$