Answer = 0.5
Given, Moles of Benzene = 4.0 moles
Moles of Toulene = 6.0 moles
Partial pressure of Benzene= $p_1 =x_1 p_1^0 = \frac{4}{4+6} \times 120\; torr = 0.4 \times 120\; torr$
Partial pressure of Toulene= $p_2 =x_1 p_1^0 = \frac{6}{4+6} \times 80\; torr = 0.6 \times 80\; torr$
According to Dalton’s law of partial pressures, the total pressure over the solution phase $p_T = p_1 + p_2 = x_1p_1^0 + x_2p_2^0$
Total pressure $p = 120 \times 0.4 + 80 \times 0.6 = 96\; torr$
At equilibrium, if $x$ be the mole fraction of Benzene in vapour phase,
Partial pressure of Benzene = Total pressure of the solution $\times$ mole fraction of Benzene in vapour pressure
Therefore, $80\; torr \times 0.6 = 96 \; torr \times x \Rightarrow x = 0.5$