JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

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 $H_{2} O$ at $2 7^{\circ} C$ . Osmotic pressure of the solution will be

[Given : R = 0.082 L atm K $^{- 1}$ mol $^{- 1}$ ]

A
1.23 atm
B
2.46 atm
C
0.82 atm
D
1.47 atm

Solution

Moles of A $= \frac{0.3}{60} = \frac{1}{200}$ mol.

Moles of B $= \frac{0.9}{180} = \frac{1}{200}$ mol.

Total moles $= \frac{2}{200} = \frac{1}{100}$ mol in 100 mL $= 0.1$ L.

Total molarity $= \frac{1/100}{0.1} = \frac{1}{10}$ M.

$π = C R T = \frac{1}{10} \times 0.082 \times 300 = 2.46$ atm.

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