$200 \mathrm{~mL}$ of an aqueous solution of a protein contains its $1.26 \mathrm{~g}$. The osmotic pressure…
$200 \mathrm{~mL}$ of an aqueous solution of a protein contains its $1.26 \mathrm{~g}$. The osmotic pressure of this solution at $300 \mathrm{~K}$ is found to be $2.57 \times 10^{-3}$ bar. The molar mass of protein will be
$51022 \mathrm{~g} \mathrm{~mol}^{-1}$
$122044 \mathrm{~g} \mathrm{~mol}^{-1}$
$31011 \mathrm{~g} \mathrm{~mol}^{-1}$
$61038 \mathrm{~g} \mathrm{~mol}^{-1}$
Solution
$\begin{aligned}
& \because \pi=C R T \\
&=\frac{w \times 1000}{M \times V(\text { in mL })} \times R T \\
& \quad\left[\because C=\frac{n \times 1000}{V(\text { in mL })} \text { and } n=\frac{w}{M}\right] \\
&=\frac{1.26 \times 1000 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 200} \\
&=61038 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}$