Answer

It is given that:

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

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

$p_{\text {total }}=600 \mathrm{~mm}$ of $\mathrm{Hg}$

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, $p_{\text {total }}=p_{\mathrm{A}}+p_{\mathrm{B}}$

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

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

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

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

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

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

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

$=1-0.4$

$=0.6$

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

$=450 \times 0.4$

$=180 \mathrm{~mm}$ of $\mathrm{Hg}$

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

$=700 \times 0.6$

$=420 \mathrm{~mm}$ of $\mathrm{Hg}$

Now, in the vapour phase:

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

$$ \begin{aligned} & =\frac{180}{180+420} \\ & =\frac{180}{600} \\ & =0.30 \end{aligned} $$

And, mole fraction of liquid $B=1-0.30$ $=0.70$