JEE Chemistry 2026 Question with Solution

Given: Osmotic pressure of the living cell is $12 \, \text{atm}$, temperature is $300 \, \text{K}$, and $R = 0.08 \, \text{L atm K}^{-1} \, \text{mol}^{-1}$.

Find: The strength of NaCl solution isotonic with the cell in $\text{g L}^{-1}$.

Isotonic solutions possess the same osmotic pressure at a given temperature. The osmotic pressure of an electrolyte solution is given by

$$\pi = iCRT$$

For NaCl, assuming complete dissociation,

$$i = 2$$

Molar mass of NaCl is

$$23 + 35.5 = 58.5 \, \text{g mol}^{-1}$$

Equating the osmotic pressure of the cell to that of the NaCl solution,

$$12 = 2 \times C \times 0.08 \times 300$$ $$12 = 48C$$ $$C = \frac{12}{48} = 0.25 \, \text{mol L}^{-1}$$

Now, strength of solution is

$$\text{Strength} = C \times M$$ $$\text{Strength} = 0.25 \times 58.5 = 14.625 \, \text{g L}^{-1}$$

Rounding to the nearest integer,

$$15$$

Therefore, the required strength is $15 \, \text{g L}^{-1}$.

Given: The solution must be isotonic with a living cell having osmotic pressure $12 \, \text{atm}$ at $300 \, \text{K}$.

Find: Concentration and then strength of NaCl solution.

The key idea is that isotonic means equal osmotic pressure at the same temperature. Since NaCl is an electrolyte, the van't Hoff factor must be included.

Using

$$\pi = iCRT$$

Substitute the given values:

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

First evaluate the constant part:

$$2 \times 0.08 \times 300 = 48$$

So,

$$12 = 48C$$

Hence,

$$C = \frac{12}{48} = 0.25 \, \text{mol L}^{-1}$$

Now calculate molar mass of NaCl:

$$M = 23 + 35.5 = 58.5 \, \text{g mol}^{-1}$$

Convert molarity to strength:

$$\text{Strength} = 0.25 \times 58.5 = 14.625 \, \text{g L}^{-1}$$

Nearest integer is $15$.

Therefore, the numerical answer is 15.