At $20°\mathrm{C}$ the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at $20°\mathrm{C}$ above equimolar mixture of benzene and methyl benzene is  _______$\times {10}^{-2}$

$P_{\text{Total }}=X_B{P}_B+X_{MB}{P}_{MB}$

$X_{MB}$ represents benzene's mole fraction in the solution phase

$X_{MB}$ represents methyl benzene's mole fraction in the solution phase

$Y_B{P}_{T\text{ otal }}=X_B{P}_B$

$Y_B$ represents the vapor phase mole fraction of benzene 

$Y_B=\frac{0.5\times 70}{0.5\times 70+0.5\times 2\overline{0}}=0.7777$

$Y_B=77.77\times {10}^{-2}$

$Y_B=78\times {10}^{-2}$

x=78

Hence, the correct answer is 78.