CBSE 2025 · Region 6 · Set 2 · Q25 · 3 marks
Vapour pressure of pure water at $\displaystyle 298$ K is $\displaystyle 24.8$ mm Hg . Calculate the lowering in vapour pressure of an aqueous solution which freezes at $\displaystyle -0 \cdot 3^{\circ} \mathrm{C} .\left(\mathrm{K}_{\mathrm{f}}\right.$ of water $\displaystyle \left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
Marking-scheme solution
$\displaystyle \Delta T_f = K_f\, m$$\displaystyle m = \Delta T_f / K_f$$\displaystyle m = 0.3/1.86$$\displaystyle = 0.16\ m$$\displaystyle m = \dfrac{x_2 \times 1000}{M_A}$$\displaystyle x_2 = \dfrac{0.16 \times 18}{1000} = 2.88 \times 10^{-3}$$\displaystyle \dfrac{p_1^{0} - p_1}{p_1^{0}} = x_2$$\displaystyle \dfrac{24.8 - p_1}{24.8} = 2.88 \times 10^{-3}$$\displaystyle p_1^{0} - p_1 = x_2\, p_1^{0}$$\displaystyle = 2.88 \times 10^{-3} \times 24.8\ \text{mm Hg}$$\displaystyle = 0.07\ \text{mm Hg}$
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.