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 \_\_\_\_ text g L^-1. (Nearest integer) Given: R = 0.08 text L atm K^-1 text mol^-1 Assume complete dissociation of textNaCl (Given: Molar mass of Na and Cl are 23 and 35.5 text g mol^-1 respectively.)

Numerical Answer Type:
Enter a numerical value Answer: 15 to 15 +4 marks

Solution & Explanation

### Related Formula pi = i C R T ### Core Logic Given: - pi = 12 text atm - T = 300 text K - i = 2 (for complete dissociation of textNaCl) - R = 0.08 text L atm K^-1 text mol^-1 12 = 2 times C times 0.08 times 300 implies 12 = 48C implies C = 0.25 text mol/L ### Step 1: Calculating Strength Molar mass of textNaCl = 23 + 35.5 = 58.5 text g/mol. Strength = 0.25 times 58.5 = 14.625 text g/L approx 15 text g/L. ### Pattern Recognition Sees: osmotic pressure calculation for isotonic solutions with electrolyte dissociation. Trap: Forgetting Van 't Hoff factor i = 2 for textNaCl. ### Chapter Mix Class 12 Chemistry: Solutions

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