200ml of an aqueous solution of a protein contains its 1.26 g. The Osmotic Pressure of this solution at 300 K is found to be 2.57x 10^-3 bar. The molar mass of protein will be ( R = 0.083L bar mol^-1 K^-1)

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61038 g $m o l^{- 1}$

122044 g $m o l^{- 1}$

31011 g $m o l^{- 1}$

51022 g $m o l^{- 1}$

answer is B.

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Detailed Solution

$\begin{array}{l} π = C R T = \frac{w t \times 1000}{G M M \times V} R T \\ 2.57 \times 10^{- 3} = \frac{1.26 \times 1000}{G M M \times 200} \times 0.083 \times 300 \\ G M M = 61038 g \end{array}$

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