Q25 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry
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.)
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.)
Correct answer: 15
Solution
For osmotic pressure, using the van 't Hoff equation: $\pi = iMRT$.
The living cell has osmotic pressure of 12 atm.
For an isotonic NaCl solution, the osmotic pressures must be equal.
For NaCl solution: $\pi = i \times M \times R \times T$, where $i = 2$ (complete dissociation into Na⁺ and Cl⁻).
$12 = 2 \times M \times 0.08 \times 300$
$12 = 48M$
$M = 0.25$ mol/L
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.
Concentration in g/L = $0.25 \times 58.5 = 14.625 \approx 15$ g/L
The living cell has osmotic pressure of 12 atm.
For an isotonic NaCl solution, the osmotic pressures must be equal.
For NaCl solution: $\pi = i \times M \times R \times T$, where $i = 2$ (complete dissociation into Na⁺ and Cl⁻).
$12 = 2 \times M \times 0.08 \times 300$
$12 = 48M$
$M = 0.25$ mol/L
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.
Concentration in g/L = $0.25 \times 58.5 = 14.625 \approx 15$ g/L
Hence, the correct answer is 15.