JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
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.)
- A 15
- B 30
- C 7.5
- D 58.5
Answer & Solution
Correct Answer
(A) 15
Step-by-step Solution
Detailed explanation
\(\pi= iCRT\) \(12=2 \times C \times 0.08 \times 300\) \(12=2 \times C \times 24\) \(C =\frac{1}{4} mole / L\) then strength of NaCl solution \(=\frac{1}{4} \times 58.5 g / L\) \(=14.625 g / L\) \(=15 g / L\)
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