JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Solutions

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) $9.96 \times 1 0^{- 4}$ ; $9.96 \times 1 0^{- 4}$ ; $9.96 \times 1 0^{- 4}$ ✓ Correct
(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) $4.98 \times 1 0^{- 4}$ ; $4.98 \times 1 0^{- 4}$ ; $4.98 \times 1 0^{- 4}$

Answer: (A)

Solution

Osmotic pressure follows

π = (n / V) R T

. For solution A,

π_{A} = \frac{1/50000}{0.5} \times 0.083 \times 300 = 9.96 \times 1 0^{- 4}

bar. For solution B,

π_{B} = \frac{2/50000}{1} \times 0.083 \times 300 = 9.96 \times 1 0^{- 4}

bar. On mixing, moles and volume add to

3/50000

mol in

1.5

L, giving the same concentration and

π = \frac{3/50000}{1.5} \times 0.083 \times 300 = 9.96 \times 1 0^{- 4}

bar. The three values x, y and z are each

9.96 \times 1 0^{- 4}

bar.

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Solutions →

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Solutions

Solution

More JEE Main questions on Solutions