Equal moles of benzene and toluene are mixed the V.P. of benzene and toluene in pure state are 700 and 600 mm Hg respectively. The mole fraction of benzene in vapour state is 

Let the total no. of moles of benzene and toluene is 1 mole.
So, $\text{ benzene }=\frac{1}{2},\text{ Toluene }=\frac{1}{2}\text{ Total }=\frac{1}{2}+\frac{1}{2}=1$
$\begin{array}{r}{\mathrm{p}}_{\text{benzene }}^{\circ }=700\\ {\mathrm{p}}_{\text{Toluene }}^{\circ }=600\end{array}$ $\mathrm{Mole} \mathrm{fraction} \mathrm{of} \mathrm{benzene} \mathrm{in} \mathrm{vapours}$
$\begin{array}{l}\mathrm{\chi }=\frac{{\mathrm{p}}_{\text{benzene }}}{{\mathrm{p}}_{\text{total }}}\\ {\mathrm{\chi }}_{\text{benzene }}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{1}{2}\\ {\mathrm{\chi }}_{\text{toluene }}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{1}{2}\end{array}$
${\mathrm{\chi }}_{\text{benzenee }\left(\mathrm{g}\right)}=\frac{{\mathrm{p}}_{\text{benzene }}^{\circ }{\mathrm{x}}_{\text{benzene }}}{{\left({\mathrm{P}}^{\circ }\mathrm{X}\right)}_{\text{benzene }}+{\left({\mathrm{p}}^{\circ }\mathrm{X}\right)}_{\text{toluene }}}$
$=\frac{700\times \frac{1}{2}}{700\times \frac{1}{2}+600\times \frac{1}{2}}=\frac{350}{350+300}=0.54$