JEE Chemistry 2026 Question with Solution

Given:

• Mass of solute A = $0.3 \, \text{g}$
• Molar mass of A = $60 \, \text{g mol}^{-1}$
• Mass of solute B = $0.9 \, \text{g}$
• Molar mass of B = $180 \, \text{g mol}^{-1}$
• Volume of solution = $0.1 \, \text{L}$
• Temperature = $300 \, \text{K}$
• Gas constant = $R = 0.082 \, \text{L atm K}^{-1} \text{ mol}^{-1}$

Find: Osmotic pressure of the solution.

For non-electrolytes, osmotic pressure depends only on the total number of solute particles present in the solution.

Step 1: Calculate moles of solute A.

$$n_A = \frac{0.3}{60} = 0.005 \text{ mol}$$

Step 2: Calculate moles of solute B.

$$n_B = \frac{0.9}{180} = 0.005 \text{ mol}$$

Step 3: Calculate total moles of solute.

$$n = 0.005 + 0.005 = 0.01 \text{ mol}$$

Step 4: Apply osmotic pressure formula.

$$\pi = \frac{nRT}{V}$$ $$\pi = \frac{0.01 \times 0.082 \times 300}{0.1}$$ $$\pi = 2.46 \text{ atm}$$

Conclusion: The osmotic pressure of the solution is $2.46 \, \text{atm}$. Therefore, the correct option is C.

Given: Both solutes are non-volatile and non-electrolytes.

Find: Which option matches the osmotic pressure.

Since both solutes are non-electrolytes, neither dissociates in water. Therefore, the effective number of solute particles is obtained by adding their moles directly.

$$\frac{0.3}{60} = 0.005$$ $$\frac{0.9}{180} = 0.005$$ $$\text{Total solute particles in moles} = 0.01$$

Now use

$$\pi = CRT = \frac{n}{V}RT$$ $$\pi = \frac{0.01}{0.1} \times 0.082 \times 300$$ $$\pi = 2.46 \text{ atm}$$

Thus, the correct option is C.