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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2025 · 28 January · Morning Shift · Chemistry · Solutions

What is the freezing point depression constant of a solvent, 50 g...

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g mol−1256 \mathrm{~g} \mathrm{~mol}^{-1} ) and the decrease in freezing point is 0.40 K ?

Options

  1. A3.72 K kg mol−13.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}
  2. B1.86 K kg mol−11.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}
  3. C4.43 K kg mol−14.43 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}
  4. D5.12 K kg mol−15.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

Correct answer

D. 5.12 K kg mol−15.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

Step-by-step solution

ΔTf=Kb⋅m0.4=Kb125650×10−3 Kb=5.12 K kg/mol\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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