JEE Main 2025 Jan 28 Shift 1, Chemistry Q15: Colligative Properties

JEE Main2025Jan 28, Shift 1Chemistry

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 $^{- 1}$ ) and the decrease in freezing point is 0.40 K?

A
5.12 K kg mol $^{- 1}$
B
4.43 K kg mol $^{- 1}$
C
1.86 K kg mol $^{- 1}$
D
3.72 K kg mol $^{- 1}$

Solution

Use $Δ T_{f} = K_{f} \cdot m$ , where molality $m$ uses kg of solvent.

$m = \frac{1/256}{50 \times 1 0 ^{- 3}}$ mol kg $^{- 1}$

$K_{f} = \frac{Δ T _{f}}{m} = \frac{0.40 \times 256 \times 50 \times 1 0 ^{- 3}}{1} = 5.12$ K kg mol $^{- 1}$ .

$∴$ $K_{f} = 5.12$ K kg mol $^{- 1}$ .

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JEE Main 2025 Jan 28 Shift 1, Chemistry Q15: Colligative Properties

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