CBSE 2024 · Region 4 · Set 3 · Q18 · 2 marks
A solution containing $\displaystyle 60$ g of a non-volatile solute in $\displaystyle 250$ g of water freezes at $\displaystyle 270.67$ K. Calculate the molar mass of the solute. ( $\displaystyle \mathrm{K}_{\mathrm{f}}$ of water = $\displaystyle 1.86$ K $\displaystyle \mathrm{kg} \mathrm{mol}^{-1}$ ).
Marking-scheme solution
\[\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{~m} \\
& \mathrm{M}_{\mathrm{B}}=\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{~W}_{\mathrm{A}} \times \Delta \mathrm{T}_{\mathrm{f}}} \\
& =273 \cdot 15-270 \cdot 67=2 \cdot 48 \mathrm{~K} \\
& \mathrm{M}_{\mathrm{B}}=\frac{1 \cdot 86 \times 60 \times 1000}{250 \times 2 \cdot 48} \\
& =180 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
\]
SolutionsColligative Properties and Determination of Molar MassApplyvery_short_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.