Vapour pressure of two pure liquids A and B are 450 mm and 700 mm Hg respectively at ${\text{623}}^0\mathrm{C}$. If the total vapour pressure of the liquid mixture is 600 mm then the mole fraction of ‘A’ in vapour phase is equal to

According to Raoult’s law

$\begin{array}{l}{\mathrm{P}}_{\mathrm{total}}={\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{B}}\\ 600=450\left({\mathrm{X}}_{\mathrm{A}}\right)+700\left(1-{\mathrm{X}}_{\mathrm{A}}\right)\\ 600=450{\mathrm{X}}_{\mathrm{A}}+700-700{\mathrm{X}}_{\mathrm{A}}\\ {\mathrm{X}}_{\mathrm{A}}=\frac{100}{250}=0.4\\ {\mathrm{X}}_{\mathrm{B}}=0.6\end{array}$

${\mathrm{X}}_{\mathrm{A}},{\mathrm{X}}_{\mathbf{B}}$ …. M.F of A and B in liquid mixture 
${\mathrm{Y}}_{\mathrm{A}},{\mathrm{Y}}_{\mathrm{B}}$…. M.F of A and B in vapour phase
In vapour phase applying Dalton’s law
${\mathrm{Y}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{total}}}=\frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{A}}}{600}=\frac{450\left(0.4\right)}{600}=0.3$