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.) [2026]

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.) [2026]

Smart Achievers · Accepted Answer

(15)

$π = i C R T$

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

$12 = 2 \times C \times 24$

$C = \frac{1}{4} mole/L$

$then strength of NaCl solution$

$= \frac{1}{4} \times 58.5 g/L$

$= 14.625 g/L$

$\approx 15 g/L$

Solution

Ans.
(15)

$π = i C R T$

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

$12 = 2 \times C \times 24$

$C = \frac{1}{4} mole/L$

$then strength of NaCl solution$

$= \frac{1}{4} \times 58.5 g/L$

$= 14.625 g/L$

$\approx 15 g/L$

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Solution

Ans. (15) π=iCRT 12=2×C×0.08×300 12=2×C×24 C=14 mole/L then strength of NaCl solution =14×58.5 g/L =14.625 g/L ≈15 g/L

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Ans.
(15)

$π = i C R T$

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

$12 = 2 \times C \times 24$

$C = \frac{1}{4} mole/L$

$then strength of NaCl solution$

$= \frac{1}{4} \times 58.5 g/L$

$= 14.625 g/L$

$\approx 15 g/L$