A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute ‘A’ of molar mass 60 g mol⁻¹ and 0.9 g of a non-volatile non-electrolyte solute ‘B’ of molar mass 180 g mol⁻¹ in 100 mL H₂O at 27°C. Osmotic pressure of the solution will be [Given: R = 0.082 L atm K⁻¹ mol⁻¹]

A solution is prepared by dissolving non volatile non electrolytes A and B in water at 27°C find osmotic pressure

Q. A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute ‘A’ of molar mass 60 g mol⁻¹ and 0.9 g of a non-volatile non-electrolyte solute ‘B’ of molar mass 180 g mol⁻¹ in 100 mL H₂O at 27°C. Osmotic pressure of the solution will be

[Given: R = 0.082 L atm K⁻¹ mol⁻¹]

(A) 2.46 atm

(B) 0.82 atm

(D) 1.23 atm

Correct Answer: 2.46 atm

Explanation (Complete Step by Step Calculation)

For dilute solutions of non-volatile non-electrolytes, osmotic pressure is given by van’t Hoff equation:

\[ \pi = \frac{nRT}{V} \]

Moles of solute A:

\[ n_A = \frac{0.3}{60} = 0.005 \text{ mol} \]

Moles of solute B:

\[ n_B = \frac{0.9}{180} = 0.005 \text{ mol} \]

Total moles of solute in solution:

\[ n = n_A + n_B = 0.005 + 0.005 = 0.01 \text{ mol} \]

Volume of solution:

\[ V = 100 \text{ mL} = 0.1 \text{ L} \]

Temperature:

\[ T = 27^\circ\text{C} = 300 \text{ K} \]

Substitute values in van’t Hoff equation:

\[ \pi = \frac{0.01 \times 0.082 \times 300}{0.1} \]

\[ \pi = \frac{0.246}{0.1} \]

\[ \pi = 2.46 \text{ atm} \]

Hence, the osmotic pressure of the solution is 2.46 atm.

Explanation (Complete Step by Step Calculation)

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