Benzene and toluene form a solution which is nearly ideal.
Let us assume a solution of benzene ($p^0=120\; torr$) and toluene($p^0=80\; torr$) having 2.0 mole of each.
Partial pressure of Benzene= $p_1 =x_1 p_1^0 = \frac{2}{2+2} \times 120\; torr = 0.5 \times 120\; torr = 60 \; torr$
Partial pressure of Toulene= $p_2 =x_1 p_1^0 = \frac{2}{2+2} \times 80\; torr = 0.5 \times 80\; torr = 40\; torr$
According to Dalton's law of partial pressures, $\therefore$ Total pressure $p = 0.5 \times (120+80) = 100\; torr$
Composition of the vapor phase can be obtained from the vapor pressures and Dalton's law of partial pressures, $X_1 = \large \frac{x_1p_1^0}{p} = \frac{60}{100} = 0.6$
$X_2 = \large \frac{x_2p_2^0}{p} = \frac{40}{100} = 0.4$
$\therefore \Delta H_{soln}=0$
The composition of the vapor is not the same as the composition of the liquid. The vapor phase is much richer in the more volatile compound, benzene.
Gibbs free energy of mixing, $\Delta G_{mix} = G_{mixed} - G_{unmixed} = nRT(X_1 ln X_1 + X_2 ln X_2 + ...)$
The Gibbs free energy of mixing is negative and thus a spontaneous process at constant temperature and pressure.
$\Delta S_{mix} = \large-(\frac{\delta \Delta G_{mix}}{\delta T})_{p_i \; n_i} = -nR(X_1 ln X_1 + X_2 ln X_2 + ...)$$
Now, $\Delta H_{mix} = \Delta G_{mix} + T \Delta S_{mix} = nRT (X_1 ln X_1 + X_2 ln X_2 + ...) + T (-nR) \;(X_1 ln X_1 + X_2 ln X_2 + ...) = 0$
Hence (a) is the correct answer.