At 20$^\circ$ 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$^\circ$ above an equimolar mixture of benzene and methyl benzene is _____________ $\times$ 10^$-$2. (Nearest integer)

Vapour pressure of pure benzene, $p_A^o$ = 70 Torr<br/><br/>Vapour pressure of pure methyl benzene, $p_B^o$ = 20 Torr<br/><br/>This mixture is equimolar, so number of moles of benzene, <br/><br/>n_A = number of moles of methyl benzene, n_B<br/><br/>Mole fraction of benzene in vapour phase, y_A = ${{{p_A}} \over {{p_T}}}$ .... (i)<br/><br/>where p_A is pressure of benzene in mixture.<br/><br/>$${p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}$$<br/><br/>p_T is total pressure,<br/><br/>${p_T} = p_A^o{\chi _A} + p_B^o{\chi _B} = {p_A} + {p_B}$ (pressure of methyl benzene)<br/><br/>${p_T} = {{p_A^o} \over 2} + {{p_B^o} \over 2} = {{p_A^o + p_B^o} \over 2}$<br/><br/>Putting in above equation (i),<br/><br/>${y_A} = {{{{70} \over 2}} \over {{{(70 + 20)} \over 2}}} = {{70} \over {90}}$<br/><br/>y_A = 0.78 = 78 $\times$ 10^$-$2