JEE MAIN - Chemistry (2025 - 28th January Morning Shift - No. 2)
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 ?
$4.43 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
$3.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
$5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
$1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
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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