Chemistry · Solutions and Colligative Properties

JEE Advanced 2023 — Paper 2 — Question 34

50 mL of 0.2 molal urea solution (density $=1.012 \mathrm{~g} \mathrm{~mL}^{-1}$ at 300 K ) is mixed with 250 mL of a solution containing 0.06 g of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in Torr) of the resulting solution at 300 K is $\_\_\_\_$ -. [Use: Molar mass of urea $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ ; gas constant, $\mathrm{R}=62 \mathrm{~L} \mathrm{Torr} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ; Assume, $\Delta_{\text {mix }} \mathrm{H}=0, \Delta_{\text {mix }} \mathrm{V}=0$ ]

Answer: 682

Numerical answer — enter this value.

Step-by-step solution

0.2 molal means 0.2 moles in 1000 g of solvent.

Volume $=\frac{M}{d}$ Mass of solution $=1012 \mathrm{~g}$ Volume $=\frac{1012 \mathrm{~g}}{1.012 \mathrm{~g} \mathrm{ml}^{-1}}$ $\mathrm{V}=1000.00 \mathrm{ml}$ $1000.00 \mathrm{ml} \longrightarrow 0.2$ moles 50 ml of solution $=\frac{0.2}{1000} \times 50$ moles $\mathrm{n}_{\text {urea }}=0.01$ moles In $2^{\text {nd }}$ solution: $\mathrm{n}_{\text {urea }}=\frac{0.06}{60}=0.001$ Final molarity $(M)=\frac{n_{1}+n_{2}}{V_{1}+V_{2}}=\frac{0.01+0.001}{\frac{(50+250)}{1000}}$ $M=\frac{11}{300}$ $\pi=$ CRT $=\frac{11}{300} \times 62 \times 300$ $=682$ torr

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)