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The vapour pressure of two liquids $P$ and $\mathrm{Q}$ are 80 and 60 torr respectively. The total vapour pressure of solution obtained by mixing 3 mole of $\mathrm{P}$ and $2 \mathrm{~mol}$ of $\mathrm{Q}$ would be:
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Verified Answer
The correct answer is:
72 torr
By Raoult's Law $\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{P}}^{\circ} \mathrm{X}_{\mathrm{P}}+\mathrm{P}_{\mathrm{Q}}^{\circ} \mathrm{X}_{\mathrm{Q}}$
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{P}}^{\circ}=80 \text { torr } \\
& \mathrm{X}_{\mathrm{P}}=\frac{3}{5} \\
& \mathrm{P}_{\mathrm{Q}}^{\circ}=60 \text { torr }
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{X}_{\mathrm{Q}} & =\frac{2}{5} \\
\mathrm{P}_{\mathrm{T}} & =80 \times \frac{3}{5}+60 \times \frac{2}{5} \\
& =48+24 \\
\mathrm{P}_{\mathrm{T}} & =72 \text { torr. }
\end{aligned}
$$
Related Theory
Solutes which dissociate or associate in a particular solution will not obey Raoult's law. Raoult's law applies to solutions containing non-volatile solute only. Raoult's law cannot be applied to concentrated solutions.
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{P}}^{\circ}=80 \text { torr } \\
& \mathrm{X}_{\mathrm{P}}=\frac{3}{5} \\
& \mathrm{P}_{\mathrm{Q}}^{\circ}=60 \text { torr }
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{X}_{\mathrm{Q}} & =\frac{2}{5} \\
\mathrm{P}_{\mathrm{T}} & =80 \times \frac{3}{5}+60 \times \frac{2}{5} \\
& =48+24 \\
\mathrm{P}_{\mathrm{T}} & =72 \text { torr. }
\end{aligned}
$$
Related Theory
Solutes which dissociate or associate in a particular solution will not obey Raoult's law. Raoult's law applies to solutions containing non-volatile solute only. Raoult's law cannot be applied to concentrated solutions.
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