The osmotic pressure of a living cell in 12 atm at 300 K . The strength of sodium chloride solution that is isotonic with the living cell at this temperature is \ \ \ \ g L -1 . (Nearest integer) Given : R =0.08 L atm K -1 mol -1 Assume complete dissociation of NaCl (Given : Molar mass of Na and Cl are 23 and 35.5 g mol -1 respectively.)

Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 21 January, Evening Shift — Question 65

The osmotic pressure of a living cell in 12 atm at 300 K . The strength of sodium chloride solution that is isotonic with the living cell at this temperature is $\_\_\_\_$ $\mathrm{g} \mathrm{L}^{-1}$ . (Nearest integer)

Given : $\mathrm{R}=0.08 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ Assume complete dissociation of NaCl (Given : Molar mass of Na and Cl are 23 and 35.5 $\mathrm{g} \mathrm{mol}^{-1}$ respectively.)

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

$\pi=\mathrm{iCRT}$ $12=2 \times \mathrm{C} \times 0.08 \times 300$ $12=2 \times \mathrm{C} \times 24$ $\mathrm{C}=\frac{1}{4} \mathrm{~mole} / \mathrm{L}$ then strength of NaCl solution $=\frac{1}{4} \times 58.5 \mathrm{~g} / \mathrm{L}$ $=14.625 \mathrm{~g} / \mathrm{L}$ $=15 \mathrm{~g} / \mathrm{L}$

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)

A substance ' X ' ( 1.5 g ) dissolved in 150 g of a solvent ' Y ' (molar mass =300 g mol^-1 ) led to an elevation of the boiling point by…

Answer: 15

Step-by-step solution

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