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JEE Advanced Chemistry Solutions 2019 JEE Advanced 2019 (Paper 1)

JEE Advanced Chemistry Question (2019) — Solution

Question

On dissolving 0.5   g of a non-volatile non-ionic solute to 39   g of benzene, its vapor pressure decreases from 650   m m   H g to 640   m m   H g . The depression of freezing point of benzene (in K ) upon addition of the solute is _______ (Given data: Molar mass and the molal freezing point depression constant of benzene are 78   g   m o l - 1 and 5.12   K   k g   m o l - 1 , respectively)

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

ΔP Ps = n solute n solvent 650 - 640 640 = 1 × 0.5 × 78 M × 39 M solute = 64  g ΔT F = k f × m ⇒ 5.12 × 0.5 × 1000 64 × 39 ΔT f = 1.0256

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