Benzene and naphthalene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.

${\mathrm{n}}_{\text{benzene }}=\frac{80\mathrm{g}}{78\mathrm{g}{\mathrm{mol}}^{-1}}=1.026\mathrm{mol}$
   $\begin{array}{l}{\mathrm{n}}_{\mathrm{naphthalene} }=\frac{100}{128}=0.781\mathrm{mol}\\ {\mathrm{x}}_{\mathrm{benzene} }=\frac{1.026}{1.026+0.781}=0.568\end{array}$
$\begin{array}{l}{\mathrm{x}}_{\mathrm{naphthalene} }=1-0.568=0.432\\ {\mathrm{p}}_{\mathrm{benzene} }={\mathrm{p}}_{\mathrm{benzene} }^{\circ }\times {\mathrm{x}}_{\mathrm{berzene} }\\ =50.71\times 0.568=28.80\mathrm{mm}\end{array}$
$\begin{array}{r}{\mathrm{p}}_{\mathrm{naphthalene} }=32.06\times 0.432=13.85\mathrm{mm}\\ {\mathrm{p}}_{\mathrm{total} }=28.80+13.85=42.65\mathrm{mm}\end{array}$
$\begin{array}{l}{\mathrm{y}}_{\text{benzene }}\text{ in vapour phase }=\frac{{\mathrm{x}}_{\text{benzene }}\times {\mathrm{p}}_{\text{benzene }}^{\circ }}{{\mathrm{p}}_{\text{total }}}\\ =\frac{0.568\times 50.71}{42.65}=0.675\end{array}$