The vapour pressures of pure benzene and toluene are 160 and $60 mmHg$ respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is

We know,

Total pressure, $P=P_A^0{x}_A+P_B^0{x}_B$

$P_A^0$ is the partial pressure of $A$ and $P_B^0$ is the partial pressure of $B.x_A$ is the mole fraction of $A$

Now, vapour pressure of pure benzene, $P_A^0=160mmHg$ and that of toluene is $P_B^0=60mmHg$

So, $P=160\times \frac{1}{2}+60\times \frac{1}{2}=80+30=110mmHg$

Now the mole fraction of benzene in vapour phase,

$P_A^0{x}_A=Y_A{P}_{total}$

$\frac{P_A^0{x}_A}{P_{total}}=Y_A$

$Y_A=\frac{80}{110}=0.73$

Therefore, option (D) is correct.