CBSE 2026 · Region 1 · Set 1 · Q24 · 3 marks
Calculate the boiling point of a solution containing $\displaystyle 0.61$ g of benzoic acid (Molar mass $\displaystyle =122 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in $\displaystyle 5$ g of $\displaystyle \mathrm{CS}_{2}$ in which it dimerises to the extent of $\displaystyle 88$%. The boiling point and $\displaystyle \mathrm{K}_{\mathrm{b}}$ of $\displaystyle \mathrm{CS}_{2}$ are $\displaystyle 46.2$ °C and $\displaystyle 2.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ respectively.
Marking-scheme solution
$\displaystyle \alpha = 88\% = 0.88$$\displaystyle \Delta T_b = i\,K_b\,m = i\,K_b\,\dfrac{W_B}{M_B} \times \dfrac{1000}{W_A}$$\displaystyle 0.88 = \dfrac{1-i}{1-\dfrac{1}{2}}$$\displaystyle i = 0.56$$\displaystyle T_b - T^\circ_b = 0.56 \times 2.3 \times \dfrac{0.61}{122} \times \dfrac{1000}{5}$$\displaystyle T_b - 46.2\,^\circ\mathrm{C} = 1.288\ \mathrm{K}$ or $\displaystyle ^\circ\mathrm{C}$$\displaystyle \therefore T_b = (46.2 + 1.288)\,^\circ\mathrm{C} = 47.488\,^\circ\mathrm{C}\ /\ 320.638\ \mathrm{K}$
SolutionsColligative Properties and Determination of Molar MassApplyshort_answerhard
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CBSE Class 12 Chemistry past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.