Question

$200 \,mL$ 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 ($R = 0.083 \, L\, bar\, mol^{-1} \, K^{-1}$):

• A. $61038 \,g\, mol^{-1}$
• B. $51022 \,g\, mol^{-1}$
• C. $122044 \,g \,mol^{-1}$
• D. $31011 \,g \,mol^{-1}$

Answer 1

Answer: (A)

$61038 \,g\, mol^{-1}$

Answer 2

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