The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in cotanct with equimolar solution of benzene and toluene is

In equimolar solution of benzene and toluene, mole fraction of benzene,

${\mathrm{X}}_{\mathrm{b}}=$ mole-fraction of toluene, ${\mathrm{X}}_{\mathrm{t}}=\frac{1}{1+1}=0.5$

So partial pressure of benzene,

${\mathrm{p}}_{\mathrm{b}}={\mathrm{X}}_{\mathrm{b}}.{\mathrm{p}}_{\mathrm{b}}^{\mathrm{o}}=0.5\times 160$ torr = 80 torr and partial pressure of toluene,

${\mathrm{p}}_{\mathrm{t}}={\mathrm{X}}_{\mathrm{t}}{\mathrm{p}}_{\mathrm{t}}^{\mathrm{o}}=0.5\times 60$ torr = 30 torr

Hence, vapour pressure of solution = 80 torr + 30 torr = 110 torr

According to Dalton's law of partial pressure,

Partial vapour pressure of toluene,

${\mathrm{p}}_{\mathrm{t}}=$ mole fraction of touene in vapour phase $\times$ total pressure $\left({\mathrm{p}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}\right)$

or mole fraction of toluene in vapour phase $=\frac{{\mathrm{p}}_{\mathrm{t}}}{{\mathrm{p}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}=\frac{30\text{ torr}}{110\text{ torr}}=0.27$