MHT CET · Chemistry · Solutions
0.2 molal aqueous solution of KCl freezes at \(-0.680^{\circ} \mathrm{C}\). Calculate van't Hoff factor for this solution. \(\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\)
- A 1.22
- B 1.32
- C 1.42
- D 1.83
Answer & Solution
Correct Answer
(D) 1.83
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{m} \mathrm{K} \)
\( \therefore 0.68 \mathrm{~K}=\mathrm{i} \times 0.2 \mathrm{~mol} \mathrm{~kg}^{-1} \times 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \)
\( \therefore \mathrm{i}=\frac{0.68}{0.2 \times 1.86}=1.827 \approx 1.83\)
\( \therefore 0.68 \mathrm{~K}=\mathrm{i} \times 0.2 \mathrm{~mol} \mathrm{~kg}^{-1} \times 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \)
\( \therefore \mathrm{i}=\frac{0.68}{0.2 \times 1.86}=1.827 \approx 1.83\)
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