The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively,

It is given that:

$p_{\mathrm{A}}^{0}=450 \mathrm{~mm}$ of $\mathrm{Hg}$

$p_{\mathrm{B}}^{0}=700 \mathrm{~mm}$ of $\mathrm{Hg}$

ptotal = 600 mm of Hg

From Raoult’s law, we have:

$p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$

$p_{\mathrm{B}}=p_{\mathrm{B}}^{0} x_{\mathrm{B}}=p_{\mathrm{B}}^{0}\left(1-x_{\mathrm{A}}\right)$ Therefore, total pressure, $p_{\text {tocal }}=p_{\mathrm{A}}+p_{\mathrm{B}}$

$\Rightarrow p_{\text {tocal }}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}+p_{\mathrm{B}}^{0}\left(1-x_{\mathrm{A}}\right)$

$\Rightarrow p_{\text {total }}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}+p_{\mathrm{B}}^{0}-p_{\mathrm{B}}^{0} x_{\mathrm{A}}$

$\Rightarrow p_{\text {total }}=\left(p_{\AA}^{0}-p_{\mathrm{B}}^{0}\right) x_{\mathrm{A}}+p_{\mathrm{B}}^{0}$

$\Rightarrow 600=(450-700) x_{\mathrm{A}}+700$

$\Rightarrow-100=-250 x_{\mathrm{A}}$

$\Rightarrow x_{\mathrm{A}}=0.4$

Therefore, $x_{\mathrm{B}}=1-x_{\mathrm{A}}$

= 1 − 0.4

= 0.6

Now, $p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$

= 700 × 0.6

= 420 mm of Hg

Now, in the vapour phase:

Mole fraction of liquid $\mathrm{A}=\frac{p_{\mathrm{A}}}{p_{\mathrm{A}}+p_{\mathrm{B}}}$

$=\frac{180}{180+420}$

$=\frac{180}{600}$

= 0.30

And, mole fraction of liquid B = 1 − 0.30

= 0.70