Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene 0.50. If 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 will be :

Given 

${\mathrm{p}}_{\mathrm{benzene}}^0=119\mathrm{torr}{\text{\hspace{0.17em};\hspace{0.17em}p}}_{\mathrm{toluene}}^0=37\mathrm{torr}$

${\mathrm{X}}_{\mathrm{toluene}}={\mathrm{X}}_{\mathrm{benzene}}=0.5$(X = M.F. in liquid phase)

According to Raoult’s Law

${\mathrm{P}}_{\mathrm{total}}\mathrm{\hspace{0.17em}}=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{p}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{A}}\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}{p}_{\mathrm{B}}^0\hspace{0.17em}\hspace{0.17em}{\mathrm{X}}_{\mathrm{B}}\mathrm{\hspace{0.17em}}$

              = 119(0.5)  +  37(0.5)
In Vapour  phase, applying Dalton’s Law
${\mathrm{p}}_{\mathrm{toluene}}\hspace{0.17em}\hspace{0.17em}=\mathrm{\hspace{0.17em}}{\mathrm{T}}_{\mathrm{total}}\hspace{0.17em}\hspace{0.17em}\times \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{y}}_{\mathrm{toluene}}\mathrm{\hspace{0.17em}}$

(Y = M.F. in vapour phase)

${\mathrm{y}}_{\mathrm{toluene}}=\frac{{\text{p}}_{\mathrm{toluene}}}{{\mathrm{P}}_{\mathrm{total}}}=\frac{37\left(0.5\right)}{119\left(0.5\right)+37\left(0.5\right)}=0.237$