As we know that:
$P_{A}^{\circ}=450 \mathrm{~mm} \text { of } \mathrm{Hg}$
$P_{B}^{\circ}=700 \mathrm{~mm}$ of $\mathrm{Hg}$
$P_{\text {total }}=600 \mathrm{~mm} \text { of } \mathrm{Hg}$
As Raoult’s law says:
$P_{A}=P_{A}^{\circ} x_{A} P_{B}=P_{B}^{\circ} x_{B}=P_{B}^{\circ}\left(1-x_{A}\right)$
Hence, total pressure, $P_{\text {total }}=P_{A}+P_{B}$
$=P_{\text {total }}=P_{A}^{\circ} x_{A}+P_{B}^{\circ}\left(1-x_{A}\right)$
$=P_{\text {total }}=P_{A}^{\circ} x_{A}+P_{B}^{\circ}-P_{B}^{\circ} x_{A}$
$\Rightarrow P_{\text {total }}=\left(P_{A}^{\circ}-P_{B}^{\circ}\right) x_{A}+P_{B}^{\circ}$
$=>600=(450-700) \mathrm{x}_{\mathrm{A}}+700$
$\Rightarrow-100=-250 \mathrm{x}_{\mathrm{A}}$
$\Rightarrow \mathrm{XA}=0.4$
Hence, $x_{B}=1-x_{A}=1-0.4=0.6$
Now, $P_{A}=P_{A}^{\circ} x_{A}$
$=450 \times 0.4=180 \mathrm{~mm}$ of $\mathrm{Hg}$
${{P}_{B}}=P_{B}^{{}^\circ }{{x}_{B}}$
$=700\times 0.6=420~\text{mm of Hg}$
Now, in the vapour phase: Mole fraction of liquid $\mathrm{A}=\frac{P_{\mathrm{A}}}{P_{A}+P_{B}}$
$=\frac{180}{180+420}$
$=\frac{180}{600}$
$=0.30$ And, mole fraction of liquid $\mathrm{B}=1-0.30=0.70$