Molar mass of benzene $(\mathrm{C}_6\mathrm{H}_6) = 78 \text{ g mol}^{-1}$ Molar mass of naphthalene $(\mathrm{C}_{10}\mathrm{H}_8) = 128 \text{ g mol}^{-1}$ ∴ Number of moles in 80 g of benzene = $\frac{80 \text{ g}}{78 \text{ g mol}^{-1}} = 1.026 \text{ mole}$ ∴ Number of moles in 100 g of naphthalene $= \frac{100 \text{ g}}{128 \text{ g mol}^{-1}}$ = 0.78125 mole ∴ In the solution, mole fraction of benzene $= \frac{1.026}{1.026+0.78125} = \frac{1.026}{1.80725}$ Mole fraction of naphthalene = 1 – 0.567= 0.433$P^0_{\text{Benzene}} = 50.71$ mm Hg, $P^0_{\text{naphthalene}} = 32.06$ mm Hg Applying Raoult’s law of vapour pressure $P_{\text{Benzene}} = x_{\text{Benzene}} \times P^0_{\text{Benzene}}$ = 0.567 × 50.71 mm Hg = 28.75 mm Hg$P_{\text{naphthalene}} = x_{\text{naphthalene}} \times P^0_{\text{naphthalene}}$ = 0.433 × 32.06 mm Hg = 13.88 mm Hg∴ From Dalton’s law, mole fraction of benzene in the vapour phase $= \frac{P_{\text{Benzene}}}{P_{\text{Benzene}}+P_{\text{naphthalene}}}$ $= \frac{28.75}{28.75+13.88} = \frac{28.75}{42.63} = 0.6744$