The vapour pressures of A and B at 25 ° C are 90   mmHg and 15   mm   Hg respectively.…

Given ${\mathrm{P}}_{\mathrm{A}}°=90 \mathrm{mmHg},$ at $25°\mathrm{C}$

${\mathrm{P}}_{\mathrm{B}}{}^{\circ }=15 \mathrm{mmHg}$

and $\left.\begin{array}{l}{\mathrm{X}}_{\mathrm{A}}=0.6\\ {\mathrm{X}}_{\mathrm{B}}=0.4\end{array}\right\}{\mathrm{P}}_{\mathrm{T}}={\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}+{\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}$

$=(0.6\times 90)+(0.4\times 15)$

$=54+6=60 \mathrm{mm}$

Now mol fraction of $\mathrm{B}$ in the vapour phase

i.e. ${\mathrm{Y}}_{\mathrm{B}}=\frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{P}}_{\mathrm{T}}}=\frac{{\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}}{60}=0.1=1\times {10}^{-1}$

therefore: $\mathrm{x}=1$