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← JEE Main 2022 · 26 Jun · Shift 2
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JEE Main 2022 · 26 Jun · Shift 2, Chemistry question 85

The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol−1^{-1} in 200 mL of water at 27°C is ______ Pa. [integer value]

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

Answer and worked solution

Official answer: 415

Concentration. 2.0/600000.2=1.67×10−4\dfrac{2.0/60000}{0.2}=1.67\times10^{-4} mol L−1^{-1}.

Osmotic pressure. π=CRT=1.67×10−4×0.083×300≈4.15×10−3\pi=CRT=1.67\times10^{-4}\times0.083\times300\approx4.15\times10^{-3} bar =415=415 Pa.

Answer: 415

Topics:Colligative properties
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