The osmotic pressure of a living cell is 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.)

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Solutions

The osmotic pressure of a living cell is 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.)

Answer: 15

Solution

Isotonic solutions share the same osmotic pressure, so for the sodium chloride solution

π = i C R T

with

i = 2

for complete dissociation. Solving

12 = 2 \times C \times 0.08 \times 300

gives

C = 0.25 mol L^{- 1}

. Multiplying by the molar mass of sodium chloride,

58.5 g mo l^{- 1}

, gives

0.25 \times 58.5 = 14.6 g L^{- 1}

. To the nearest integer the required strength is 15 g L

^{- 1}

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer

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