0.2 molal aqueous solution of KCl freezes at $-0.680^{\circ} \mathrm{C}$. Calculate van't Hoff factor for…

0.2 molal aqueous solution of KCl freezes at $-0.680^{\circ} \mathrm{C}$. Calculate van't Hoff factor for this solution. $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
  1. 1.22
  2. 1.32
  3. 1.42
  4. 1.83

Solution

$\begin{array}{ll} & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{m} \mathrm{K} \\ \therefore & 0.68 \mathrm{~K}=\mathrm{i} \times 0.2 \mathrm{~mol} \mathrm{~kg}^{-1} \times 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\ \therefore \quad & \mathrm{i}=\frac{0.68}{0.2 \times 1.86}=1.827 \approx 1.83\end{array}$

Asked in: MHT CET 2024 (10 May Shift 1)

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