CBSE 2024 · Region 2 · Set 2 · Q24 · 3 marks
The vapour pressure of a solvent at $\displaystyle 283$ K is $\displaystyle 100$ mm Hg . Calculate the vapour pressure of a dilute solution containing $\displaystyle 1$ mole of a strong electrolyte AB in $\displaystyle 50$ moles of the solvent at $\displaystyle 283$ K (assuming complete dissociation of solute AB).
Marking-scheme solution
\[\begin{aligned}
& \frac{p^{\circ}-p_{\mathrm{s}}}{\mathrm{P}^{\circ}}=i \times \chi \\
& A B \longrightarrow A^{+}+B^{-} \\
& \quad i=2
\end{aligned}
\]
\(\displaystyle \frac{100-\mathrm{P}_{S}}{100}=2 \times \frac{n_{A B}}{n_{\text {solvent }}}\) (For dilute solution)
\[\begin{gathered}
\frac{100-\mathrm{P}_{\mathrm{s}}}{100}=2 \frac{1}{50} \\
100-\mathrm{P}_{\mathrm{s}}=4
\end{gathered}
\]
\(\displaystyle \mathrm{P}_{\mathrm{s}}=96 \mathrm{~mm} \mathrm{Hg}\)
SolutionsVapour Pressure of Liquid SolutionsApplyshort_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.