Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 5 April, Evening Shift — Question 68

20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water.

The molar mass of hemoglobin is $\_\_\_\_$ $\mathrm{kg} \mathrm{mol}^{-1}$ . (Nearest integer) (Given : $\mathrm{g}=10 \mathrm{~ms}^{-2}, \mathrm{R}=8.3 \mathrm{kPa} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ , density of solution $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ )

Answer: 62

Numerical answer — enter this value.

Step-by-step solution

Osmotic pressure, $\pi=\rho \mathrm{gh}$ $=1000 \times 10 \times 80 \times 10^{-3}$

=800 \mathrm{~Pa} \end{gathered}$$ Let molar mass of haemoglobin $=\mathrm{M} \mathrm{g} / \mathrm{mol}$ conc. of haemoglobin $=\frac{20 / \mathrm{M}}{10^{-3}}\left(\frac{\mathrm{~mol}}{\mathrm{~m}^{3}}\right)$ $\pi=$ CRT (S.I. units) $\Rightarrow \pi=\left[\frac{20}{10^{-3} \mathrm{M}}\right] \times 8.3 \times 300$ $=800$[0pt] [From ] Solving we get : $\mathrm{M}=\frac{20 \times 8.3 \times 300}{0.8} \mathrm{~g} / \mathrm{mol}$ $=62.25 \mathrm{~kg} \mathrm{~mol}^{-1}$ Ans $=62$

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)