JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Solutions

An aqueous solution of volume

300 cm^{3}

contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is ______

g mol^{- 1}

Given:

R = 0.083 L

bar

K^{- 1} mol^{- 1}

Answer: 40535

Solution

From the osmotic-pressure relation

π = \frac{w}{M V} R T

the molar mass is

M = \frac{w R T}{π V}

. Using

w = 0.63 g

R = 0.083 L bar K^{- 1} mo l^{- 1}

T = 300 K

π = 1.29 \times 1 0^{- 3} bar

and

V = 0.3 L

M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 1 0 ^{- 3} \times 0.3} = \frac{15.687}{3.87 \times 1 0 ^{- 4}}

. The molar mass of the protein is

40535 g mo l^{- 1}

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer

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