$200c{m}^3$ of an aqueous solution of a protein contains $1.26g$ of the protein. The osmotic pressure of such a solution at $300K$ is found to be $2.57\times {10}^{-3}bar$. Calculate the molar mass of the protein.

Given:
Osmotic pressure $\left(\pi \right)=2.57\times {10}^{-3}bar$,
Volume of solution $\left(V\right)=200c{m}^3=0.2litre$,
Temperature $\left(T\right)=300K$
And mass of solute (protein) $=1.26g$

We know that universal gas constant $\left(R\right)=0.083Lbarmo{l}^{-1}{K}^{-1}$

We know that osmotic pressure $\left(\pi \right)=CRT$
And we also know that molarity $\left(C\right)=\frac{\text{mass of solute(m\_2)}}{\text{molar mass of solute(M\_2)}\times \text{volume of solution(L)}}$
So, $\pi =\frac{\text{mass of solute(m\_2)}\times R\times T}{\text{molar mass of solute(M\_2)}\times \text{volume of solution(L)}}$

Molar mass of solute $\left(M_2\right)=\frac{1.26g\times 0.083L{K}^{-1}mo{l}^{-1}\times 300K}{2.57\times {10}^{-3}bar\times 0.2L}$

$=61,038.91gmo{l}^{-1}$