The vapour pressure of pure liquids \(A\) and \(B\) are 450 and \(700 \mathrm{~mm}\) of \(\mathrm{Hg}\) at \(350 \mathrm{~K}\) respectively. If the total vapour pressure of the mixture is \(600 \mathrm{~mm}\) of \(\mathrm{Hg}\), the composition of the mixture in the solution is

Given, vapour pressure of pure liquid \(A\), \(p_{A}^{\circ}=450 \mathrm{~mm}\) of \(\mathrm{Hg}\)
Vapour pressure of pure liquid \(B\),
\(p_{B}^{\circ}=700 \mathrm{~mm} \text { of } \mathrm{Hg}\)
Total vapour pressure, \(p_{\text {Total }}=600 \mathrm{~mm}\) of \(\mathrm{Hg}\) From Raoult's law, \(p_{\text {Total }}=p_{A}^{\circ} \chi_{A}+p_{B} \chi_{B}\)
\(\begin{aligned}
&-p_{A}^{\circ} \chi_{A}+p_{B}^{\circ}\left(1-\chi_{A}\right) \\
\Rightarrow \quad 600 &=450 \chi_{A}=700\left(1-\chi_{A}\right) \\
&=450 \chi_{A}+700-700 \chi_{A} \\
&=700-250 \chi_{A} \\
250 \chi_{A} &=700-600 \Rightarrow \chi_{A}=\frac{100}{250} \\
\therefore \quad \chi_{A} &=0.4 \\
\chi_{B} &=1-0.4=0.6
\end{aligned}\)