When 1 mole of benzene is mixed with 1 mole of toluene the vapour will contain : (Given: vapour of benzene = 12.8 kPa and vapour pressure of toluene : 3.85 kPa).

When 1 mole of benzene is mixed with 1 mole of toluene the vapour will contain higher percentage of benzene. As it is an ideal solution, it follows Raoult's Iaw. 

The vapour pressure of the solution depends on the mole fraction of the solvent.

$P_{\text{soln }}={\chi }_{\text{solvent }}{P}_{\text{solvent }}^0$

P_soln is the vapour pressure of the pure solvent

Since the mole fraction of both the components is same, but the vapour pressure of benzene is higher than the toluene, its percentage will be greater in the vapour of the solution.