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JEE Main Chemistry Solutions 2026 JEE Main 2026 (21 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

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.)

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

For osmotic pressure, using the van 't Hoff equation: = iMRT. The living cell has osmotic pressure of 12 atm. For an isotonic NaCl solution, the osmotic pressures must be equal. For NaCl solution: = i M R T, where i = 2 (complete dissociation into Na⁺ and Cl⁻). 12 = 2 M 0.08 300 12 = 48M M = 0.25 mol/L Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol. Concentration in g/L = 0.25 58.5 = 14.625 15 g/L

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