JEE Main 2026 · 2 Apr · Shift 2 · Chemistry Q55

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ChemistrySingle correct+4 / -1Question 55 of 75

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 $^{-1}$ )

9.96\times10^{-4}

;

9.96\times10^{-4}

;

9.96\times10^{-4}

9.96\times10^{-4}

;

9.96\times10^{-4}

;

19.92\times10^{-4}

4.98\times10^{-4}

;

4.98\times10^{-4}

;

9.96\times10^{-4}

4.98\times10^{-4}

;

4.98\times10^{-4}

;

4.98\times10^{-4}

Answer and worked solution

Official answer: A

Concentrations (molar mass 50000):

A: $\frac{1/50000}{0.5}=4\times10^{-5}$ M.
B: $\frac{2/50000}1=4\times10^{-5}$ M.
Mixture: $\frac{6\times10^{-5}}{1.5}=4\times10^{-5}$ M.

Osmotic pressure is the same for all three: $\pi=cRT=4\times10^{-5}\times0.083\times300=9.96\times10^{-4}$ bar.

(A)

Topics:Colligative properties

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JEE Main 2026 · 2 Apr · Shift 2, Chemistry question 55