JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass \(256 \mathrm{~g} \mathrm{~mol}^{-1}\) ) and the decrease in freezing point is 0.40 K ?
- A \(3.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- B \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- C \(4.43 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- D \(5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \Delta \mathrm{T}_{\mathrm{f}} & =\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m} \\ 0.4 & =\mathrm{K}_{\mathrm{b}} \frac{\frac{1}{256}}{50 \times 10^{-3}} \\ \mathrm{~K}_{\mathrm{b}} & =5.12 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\end{aligned}\)
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