Chemistry · Solutions and Colligative Properties

JEE Advanced 2019 — Paper 1 — Question 25

On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm Hg to 640 mm Hg . 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 $78 \mathrm{~g} \mathrm{~mol}^{-1}$ and $5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ , respectively)

Answer: 1.02

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mathrm{p}_{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{s}}}=\mathrm{i}\left(\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solvent }}}\right)$

$\frac{650-640}{640}=1 \times \frac{0.5 \times 78}{\mathrm{M} \times 39}$

$\Rightarrow \mathrm{M}_{\text {solute }}=64 \mathrm{~g}$

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times$ molality $=5.12 \times \frac{0.5 \times 1000}{64 \times 39}$

$\Delta \mathrm{T}_{\mathrm{f}}=1.02$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)