1. Tardigrade
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  3. Chemistry
  4. 200 mL 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 2.57 × 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1):

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Q. $200 \,mL$ 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 $2.57 \times 10^{-3}$ bar.
The molar mass of protein will be ($R = 0.083 \, L\, bar\, mol^{-1} \, K^{-1}$):

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Solution:

$\pi v = \frac{w}{m}RT $
$ 2.57 \times10^{-3} \times\frac{200}{1000} = \frac{1.26}{m} \times0.083 \times300 $
$ m = 61038 \; gm \; mol^{-1} $