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

(N/A) The osmotic pressure formula is given by $\Pi V = nRT = \frac{w_2}{M_2} RT$,where $M_2$ is the molar mass of the solute.
Given values:
$\Pi = 2.57 \times 10^{-3} \, bar$
$V = 200 \, cm^3 = 0.200 \, L$
$w_2 = 1.26 \, g$
$T = 300 \, K$
$R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$
Rearranging the formula for $M_2$:
$M_2 = \frac{w_2 RT}{\Pi V}$
Substituting the values:
$M_2 = \frac{1.26 \, g \times 0.083 \, L \, bar \, K^{-1} \, mol^{-1} \times 300 \, K}{2.57 \times 10^{-3} \, bar \times 0.200 \, L}$
$M_2 = \frac{31.374}{0.000514} \, g \, mol^{-1} \approx 61038.9 \, g \, mol^{-1}$