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.)

Hint:

Isotonic means $\pi_{1} = \pi_{2}$. Always check if the solute is an electrolyte to include the van't Hoff factor $i$. For calculations involving $R=0.0821$, using $1/12$ can often simplify fractions, but here $R=0.08$ was specifically given.

Answer 1

Correct Answer: 15

The osmotic pressure (π) of a solution is given by the formula: 
π = iCRT
where i is the van't Hoff factor, C is the molar concentration, R is the ideal gas constant, and T is the temperature in Kelvin.
Since NaCl dissociates into two ions (Na⁺ and Cl⁻), the van't Hoff factor \(i = 2\).
Given: π = 12 atm, T = 300 K, R = 0.08 L atm K⁻¹ mol⁻¹.

We need to find the molar concentration C:
C = π / (iRT) = 12 / (2 × 0.08 × 300) = 0.25 mol L⁻¹.

Now, calculate the molar mass of NaCl:
Molar mass of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g mol⁻¹.

To find the strength of the solution in g L⁻¹:
Strength = C × Molar mass = 0.25 mol L⁻¹ × 58.5 g mol⁻¹ = 14.625 g L⁻¹.

Rounding to the nearest integer, the strength is 15 g L⁻¹, which lies within the expected range (15,15).