(d)
Mixture contains 78 g benzene = 1 mole benzene and 46 g toluene = 0.5 mole toluene
Total mole of benzene and toluene=1.5 mol
Mole fraction of benzene in mixture
=$\frac{1}{1.5} = \frac{2}{3}$
VP of benzene $p_{b}^{\circ}$ =75 torr
$\therefore$ Partial vapour pressure of benzene = $p_{b}^{\circ}\ X_{b}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 75\ \times \frac{2}{3} = 50\ torr\$