Volume of 1 mol of benzene $$ = {{78} \over {0.877}} = 88.94$$ mL and that for 1 mol of toluene $$ = {{92} \over {0.867}} = 106.11$$ mL

At 20$$^\circ$$C, volume of 1 mol of benzene vapour $$ = 88.94 \times 2750 = 244585$$ mL = 244.58 L & that for 1 mol of toluene vapour $$ = 106.11 \times 7720 = 819.16$$ L

The vapour pressure for pure benzene, $$p_B^0 = {{nRT} \over V} = {{1 \times 0.0821 \times 293} \over {244.58}}$$ atm = 0.098 atm and that for toluene, $$p_T^0 = {{nRT} \over V} = {{1 \times 0.0821 \times 293} \over {819.16}}$$ atm = 0.029 atm

Again, total vapour pressure = VP of benzene $$\times$$ mole fraction of benzene + VP of toluene $$\times$$ mole fraction of toluene

or, $$P = p_B^0 \times {x_B} + p_T^0 \times {x_T}$$ ...... [1]

Since, $${x_B} + {x_T} = 1$$ $$\therefore$$ $${x_T} = 1 - {x_B}$$ [x_B, x_T are mole fractions of benzene & toluene, respectively]

Given, total vapour pressure = 46 torr $$ = {{46} \over {760}}$$ atm = 0.060 atm

From equation [1], we get, $$0.060 = 0.098 \times {x_B} + 0.029(1 - {x_B})$$

or, $$0.060 = 0.098{x_B} - 0.029{x_B} + 0.029$$

or, $${x_B} = 0.45$$ (in liquid phase)

$$\therefore$$ $${x_T} = 1 - 0.45 = 0.55$$ (in liquid phase)

$$\therefore$$ Mole fraction of benzene in vapour phase $$ = {{p_B^0 \times {x_B}} \over P} = {{0.098 \times 0.45} \over {0.060}} = 0.735$$