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Osmotic Pressure

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Question

A solution is prepared by dissolving $1 g$ of albumin (a protein obtained from blood plasma), in $50 c m^{3}$ of aqueous solution. The solution has an osmotic pressure of $7.6 m m H g at 300 K$ Calculate the molar mass of albumin

64.1 k g m o l^{- 1}

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88.6 k g m o l^{- 1}

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24.8 k g m o l^{- 1}

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49.3 k g m o l^{- 1}

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Solution

The correct option is D $49.3 k g m o l^{- 1}$
Osmotic pressure:

$π = \frac{n}{V} \times R T$

Here n is number of moles

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

$V = 50 c m^{3} = 0.05 L R = 0.0821 L a t m m o l^{- 1} K^{- 1} π = \frac{7.6}{760} a t m = 0.01 a t m$

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

$m_{B} = \frac{1 g \times 0.0821 L a t m m o l^{- 1} K^{- 1} \times 300 K}{0.01 a t m \times 0.05 L} m_{B} = 49260 g m o l^{- 1} m_{B} = 49.26 k g m o l^{- 1}$

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10 g

100 c m^{3}

76 m m H g at 300 K

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