An aqueous solution of volume $300 {\mathrm{cm}}^3$ contains $0.63 \mathrm{g}$ of protein. The osmotic pressure of the solution at $300 \mathrm{K}$ is $1.29 \mathrm{mbar}$. The molar mass of the protein is ${\mathrm{gmol}}^{-1}$.

Given : $\mathrm{R}=0.083 {\mathrm{L}}_{\mathrm{bar}}{\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$

The osmotic pressure of a non-electrolytic solution can be calculated as follows,

$\mathrm{\pi }=\mathrm{CRT}$

$\mathrm{\pi }$= osmotic pressure

$\mathrm{C}$ = molarity$=\frac{\mathrm{Number} \mathrm{of} \mathrm{moles}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{L}}$

$\left(1.29\times {10}^{-3}\right)=\frac{0.63}{\left(\mathrm{MW}\right)(0.3)}\times (0.083)\times 300$

$\mathrm{MW}=\frac{0.63\times 0.083\times 300}{(0.3)\times \left(1.29\times {10}^{-3}\right)}$

$= 40534.88$

$\approx 40535$ (Nearest integer)