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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
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The correct answer is:
$\chi_{A}=0.4, \chi_{B}=0.6$
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, $p_{\text {Total }}=p_{A}^{\circ} \chi_{A}+p_{B} \chi_{B}$
$\begin{aligned}
&-p_{A}^{\circ} \chi_{A}+p_{B}^{\circ}\left(1-\chi_{A}\right) \\
\Rightarrow \quad 600 &=450 \chi_{A}=700\left(1-\chi_{A}\right) \\
&=450 \chi_{A}+700-700 \chi_{A} \\
&=700-250 \chi_{A} \\
250 \chi_{A} &=700-600 \Rightarrow \chi_{A}=\frac{100}{250} \\
\therefore \quad \chi_{A} &=0.4 \\
\chi_{B} &=1-0.4=0.6
\end{aligned}$
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, $p_{\text {Total }}=p_{A}^{\circ} \chi_{A}+p_{B} \chi_{B}$
$\begin{aligned}
&-p_{A}^{\circ} \chi_{A}+p_{B}^{\circ}\left(1-\chi_{A}\right) \\
\Rightarrow \quad 600 &=450 \chi_{A}=700\left(1-\chi_{A}\right) \\
&=450 \chi_{A}+700-700 \chi_{A} \\
&=700-250 \chi_{A} \\
250 \chi_{A} &=700-600 \Rightarrow \chi_{A}=\frac{100}{250} \\
\therefore \quad \chi_{A} &=0.4 \\
\chi_{B} &=1-0.4=0.6
\end{aligned}$
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