The osmotic pressure of a dilute solution is 7 × 10 5 Pa at 273 K . Osmotic pressure of the same solution at 283 K is × 10 4 Nm -2 .

Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 29 January, Shift 1 — Question 78

The osmotic pressure of a dilute solution is $7 \times 10^{5} \mathrm{~Pa}$ at 273 K . Osmotic pressure of the same solution at 283 K is $\qquad$ $\times 10^{4} \mathrm{Nm}^{-2}$ .

Answer: 72.56

Numerical answer — enter this value.

Step-by-step solution

$\pi=$ CRT

$\Rightarrow \frac{\pi_{1}}{\pi_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$

$\Rightarrow \pi_{2}=\frac{\pi_{1} \mathrm{~T}_{2}}{\mathrm{~T}_{1}}=\frac{7 \times 10^{5} \times 283}{273}$

$=72.56 \times 10^{4} \mathrm{Nm}^{-2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)