JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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 $^{-1})$

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}$
B
$9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 19.92 \times 10^{-4}$
C
$4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 9.96 \times 10^{-4}$
D
$4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 4.98 \times 10^{-4}$

Did you get this right?

Solution

Osmotic pressure is given by $\pi = CRT = \dfrac{w}{MV}RT$

For solution A:

$x = \dfrac{1}{50000 \times 0.5} \times 0.083 \times 300$

$x = \dfrac{1}{25000} \times 24.9 = 9.96 \times 10^{-4}$ bar

For solution B:

$y = \dfrac{2}{50000 \times 1} \times 0.083 \times 300$

$y = \dfrac{2}{50000} \times 24.9 = 9.96 \times 10^{-4}$ bar

For the resultant solution (mixture of A and B):

Total mass of protein $= 1 + 2 = 3$ g

Total volume of solution $= 0.5 + 1 = 1.5$ L

$z = \dfrac{3}{50000 \times 1.5} \times 0.083 \times 300$

$z = \dfrac{3}{75000} \times 24.9 = 9.96 \times 10^{-4}$ bar

Therefore, $x = y = z = 9.96 \times 10^{-4}$ bar.

Answer: $9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}$

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Options

$9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}$

$9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 19.92 \times 10^{-4}$

$4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 9.96 \times 10^{-4}$

$4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 4.98 \times 10^{-4}$

Solution

Osmotic pressure is given by $\pi = CRT = \dfrac{w}{MV}RT$

For solution A:

$x = \dfrac{1}{50000 \times 0.5} \times 0.083 \times 300$

$x = \dfrac{1}{25000} \times 24.9 = 9.96 \times 10^{-4}$ bar

For solution B:

$y = \dfrac{2}{50000 \times 1} \times 0.083 \times 300$

$y = \dfrac{2}{50000} \times 24.9 = 9.96 \times 10^{-4}$ bar

For the resultant solution (mixture of A and B):

Total mass of protein $= 1 + 2 = 3$ g

Total volume of solution $= 0.5 + 1 = 1.5$ L

$z = \dfrac{3}{50000 \times 1.5} \times 0.083 \times 300$

$z = \dfrac{3}{75000} \times 24.9 = 9.96 \times 10^{-4}$ bar

Therefore, $x = y = z = 9.96 \times 10^{-4}$ bar.

Answer: $9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}$

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry