CBSE 2023 · Region 2 · Set 2 · Q27 · 3 marks
When $\displaystyle 19.5$ g of $\displaystyle \mathrm{F}-\mathrm{CH}_{2}-\mathrm{COOH}\left(\right.$ Molar mass $\displaystyle \left.=78 \mathrm{~g} \mathrm{~mol}^{-1}\right)$, is dissolved in $\displaystyle 500$ g of water, the depression in freezing point is observed to be $\displaystyle 1$°C. Calculate the degree of dissociation of $\displaystyle \mathrm{F}-\mathrm{CH}_{2}-\mathrm{COOH}$. [Given : $\displaystyle \mathrm{K}_{\mathrm{f}}$ for water $\displaystyle =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ]
Marking-scheme solution
\[\begin{aligned}
\Delta T_{\mathrm{f}} & =i \mathrm{K}_{\mathrm{f}} \cdot \mathrm{~m} \\
& =i \mathrm{K}_{\mathrm{f}} \frac{W_{\mathrm{B}}}{M_{\mathrm{B}}} \times \frac{1000}{W_{\mathrm{A}}} \\
1 & =i \times 1.86 \times \frac{19 \cdot 5}{78} \times \frac{1000}{500} \\
& i=1.075 \\
& \alpha=\frac{i-1}{n-1} \\
& =\frac{1.075-1}{2-1}=0.075
\end{aligned}
\]
SolutionsColligative Properties and Determination of Molar MassApplyshort_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.