CBSE 2024 · Region 3 · Set 1 · Q22 · 3 marks
A solution is prepared by dissolving $\displaystyle 5$ g of a non-volatile solute in $\displaystyle 200$ g of water. It has a vapour pressure of $\displaystyle 31.84$ mm Hg at $\displaystyle 300$ K. Calculate the molar mass of the solute. $\displaystyle 3$ (Vapour pressure of pure water at $\displaystyle 300 \mathrm{~K}=32 \mathrm{~mm} \mathrm{Hg}$ )
Marking-scheme solution
\[\begin{aligned}
& \frac{p_{1}^{o}-p_{1}}{p_{1}^{o}}=\frac{\mathrm{w}_{2} \times \mathrm{M}_{1}}{\mathrm{M}_{2} \times \mathrm{w}_{1}} \\
& \frac{(32-31.84)}{32}=\frac{5 \mathrm{~g}}{\mathrm{M}_{2}} \times \frac{18}{200 \mathrm{~g}}
\end{aligned}
\]
SolutionsColligative Properties and Determination of Molar MassApplyshort_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.