A solution is made by mixing one mole of volatile liquid $A$ with 3 moles of volatile liquid $B$. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure $B$ and the least volatile component of the solution, respectively, are :

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{P}_{\mathrm{S}}=\mathrm{P}_{\mathrm{A}}^{\mathrm{o}} \cdot \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \cdot \mathrm{X}_{\mathrm{B}} \ & 500=200 \times \frac{1}{4}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \cdot \frac{3}{4} \ & \mathrm{P}_{\mathrm{B}}^{\mathrm{o}}=600 \mathrm{~mm} \mathrm{Hg} \end{aligned}$ As $\mathrm{P}_{\mathrm{A}}^{\mathrm{o}} \lt \mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \Rightarrow \mathrm{A}$ is least volatile.

A solution is made by mixing one mole of volatile liquid \(A\) with 3 moles of volatile liquid \(B\). The…

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