A solution prepared by dissolving $10 g$ of a protein in $100 c m^{3}$ of aqueous solution has an osmotic pressure of $76 m m H g at 300 K$ Calculate the molar mass of the protein in $k g m o l^{- 1}$

24.63

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49.26

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12.32

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50.44

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Solution

The correct option is A $24.63$
Osmotic pressure:

$π = \frac{n}{V} \times R T$

Here n is number of moles

$n = \frac{Given mass}{Molar mass} = \frac{w_{B}}{m_{B}}$

$V = 100 c m^{3} = 0.1 L R = 0.0821 L a t m m o l^{- 1} K^{- 1} π = \frac{76}{760} a t m = 0.1 a t m$

$\Rightarrow π V = \frac{w_{B}}{m_{B}} \times R T m_{B} = \frac{w_{B} \times R \times T}{π \times V}$

$m_{B} = \frac{10 g \times 0.0821 L a t m m o l^{- 1} K^{- 1} \times 300 K}{0.1 a t m \times 0.1 L} m_{B} = 24630 g m o l^{- 1} m_{B} = 24.63 k g m o l^{- 1}$

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