JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

Solution A is prepared by dissolving $1$ g of a protein (molar mass $= 50000$ g mol $^{- 1}$ ) in $0.5$ L of water at $300$ K. Its osmotic pressure is $x$ bar. Solution B is made by dissolving $2$ g of same protein in $1$ L of water at $300$ K. Osmotic pressure of solution B is $y$ bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is $z$ bar. $x$ , $y$ and $z$ respectively are :

( $R = 0.083$ L bar mol $^{- 1}$ K $^{- 1}$ )

A
$9.96 \times 1 0^{- 4}$ ; $9.96 \times 1 0^{- 4}$ ; $9.96 \times 1 0^{- 4}$
B
$9.96 \times 1 0^{- 4}$ ; $9.96 \times 1 0^{- 4}$ ; $19.92 \times 1 0^{- 4}$
C
$4.98 \times 1 0^{- 4}$ ; $4.98 \times 1 0^{- 4}$ ; $9.96 \times 1 0^{- 4}$
D
$4.98 \times 1 0^{- 4}$ ; $4.98 \times 1 0^{- 4}$ ; $4.98 \times 1 0^{- 4}$

Solution

Osmotic pressure $π = C R T$ where $C$ is molar concentration.

Solution A: $C_{A} = \frac{1}{50000 \times 0.5}$ mol/L. $x = C_{A} \cdot R \cdot 300 = \frac{300 \cdot 0.083}{25000} \approx 9.96 \times 1 0^{- 4}$ bar.

Solution B: $C_{B} = \frac{2}{50000 \times 1} = \frac{1}{25000}$ mol/L. So $y = 9.96 \times 1 0^{- 4}$ bar (identical concentration).

Mixing equal-concentration solutions gives the same concentration: $z = 9.96 \times 1 0^{- 4}$ bar.

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