200 mL of an aqueous solution of a protein contains 1.26g of solute. The osmotic pressure of this solution at 300K is found to be $2.57 \times 10^{- 3}$ bar. The molar mass of protein will be $(R = 0.083 L b a r m o l^{- 1} K^{- 1})$

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

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

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

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

answer is B.

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Detailed Solution

$π = C R T$

$\begin{array}{l} = \frac{n}{v} R T \\ = \frac{W}{M} \frac{1}{V} R T \\ ∴ M = \frac{W}{π} \frac{R T}{V} \\ M = m o l a r m a s s o f p r o t e i n \\ W = w t o f p r o t e i n = 1.26 g \\ π = o s m o t i c p r e s s u r e = 2.57 \times 10^{- 3} b a r \\ V = v o l o f s o l = 200 m l = 0.2 l i t r e s \end{array}$

$\begin{array}{l} ∴ M = \frac{1.26 \times 0.083 \times 300 K}{2.57 \times 10^{- 3} \times 0.2} b y s o l v i n g \\ W e g e t \\ M = 61038 g m o l^{- 1} \end{array}$