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

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

${P}_{B} = ∘ 15mmHg$

and \left$$\begin{array}{ll}{X}_{A} = 0 . 6 \\ {X}_{B} = 0 . 4\end{array}$$\right\}{P}_{T} = {X}_{A}{P}_{A}^{o} + {X}_{B}{P}_{B}^{o}$</p><p>$= ( 0 . 6 \times 90 ) + ( 0 . 4 \times 15 )$</p><p>$= 54 + 6 = 60mm$</p><p>Now mol fraction of$B$in the vapour phase</p><p>i.e.${Y}_{B} = \dfrac{{P}_{B}}{{P}_{T}} = \dfrac{{X}_{B}{P}_{B}^{o}}{60} = 0 . 1 = 1 \times {10}^{- 1}$</p><p>therefore:$x = 1$