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JEE Main levelSingle correct optionDifficulty 2 of 3so084

A solution of $0.50 g$ of a protein in $200 mL$ of solution has an osmotic pressure of $2.00 torr$ at $300 K$ . Using $R = 62.36 L torr mo l^{- 1} K^{- 1}$ , the molar mass of the protein (in $g mo l^{- 1}$ ) is about

A $2.3 \times 1 0^{4}$
B $1.2 \times 1 0^{4}$
C $2.3 \times 1 0^{5}$
D $4.7 \times 1 0^{4}$

Answer: A

M = \frac{wRT}{πV} = \frac{0.50 \times 62.36 \times 300}{2.00 \times 0.200} = 23385 g mo l^{- 1}

, i.e. about

2.3 \times 1 0^{4}

.

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