JEE Main 2026 — Jan 21, Shift 2 — Chemistry solution

JEE Main2026Jan 21, Shift 2Chemistry

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.)

Your answer

Solution

$π = i C R T$ with $i = 2$ for fully dissociated NaCl. So $12 = 2 \cdot C \cdot 0.08 \cdot 300 = 48 C$ , giving $C = \frac{1}{4}$ mol/L.

Molar mass of NaCl $= 23 + 35.5 = 58.5$ g/mol. Strength $= \frac{1}{4} \times 58.5 = 14.625$ g/L $\approx 15$ g/L.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →