Answer

It is given that:

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

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

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

From Raoult’s law, we have: $ \quad p_{{A}}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}$

$p_{{B}}=p_{{B}}^{0} \hspace{0.8mm} \chi_{{B}}=p_{{B}}^{0}\left(1-\chi_{{A}}\right)$

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

$\Rightarrow p_{\text {total }}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}+p_{{B}}^{0}\left(1-\chi_{{A}}\right)$

$\Rightarrow p_{\text {total }}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}+p_{{B}}^{0}-p_{{B}}^{0} \chi_{{A}}$

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

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

$\Rightarrow-100=-250 \chi_{{A}}$

$\Rightarrow \chi_{{A}}=0.4$

Therefore, $\chi_{{B}}=1-\chi_{{A}}$

$\quad\quad\quad\quad\quad=1-0.4$

$\quad\quad\quad\quad\quad=0.6$

Now, $p_{{A}}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}$

$\quad\quad\quad=450 \times 0.4$

$\quad\quad\quad=180 {~mm}$ ${Hg}$

$p_{{B}}=p_{{B}}^{0} \hspace{0.8mm} \chi_{{B}}$

$\quad=700 \times 0.6$

$\quad=420 {~mm}$ ${Hg}$

Now, in the vapour phase:

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

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

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