Q.$200\ \text{mL}$ of an aqueous solution of a protein contains $1.26\ \text{g}$ of the protein. The osmotic pressure of this solution at $300\ \text{K}$ is found to be $2.57 \times 10^{-3}\ \text{atm}$. Calculate the molar mass of the protein. ($R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$.)

[!TLDR] Find $C$ from $\pi=CRT$, then moles from $C \times V$, then molar mass from mass/moles. [!ANSWER] The molar mass of the p …

From $\pi = CRT$, the molar concentration is $C = \pi/(RT) = (2.57\times10^{-3})/(0.0821 \times 300) = (2.57\times10^{-3})/24.63 \approx 1.043\times10^{-4}\ \text{mol L}^{-1}$. The moles of protein present in $200\ \text{mL} = 0.200\ \text{L}$ are $n_2 = C \times V \approx 1.043\times10^{-4} \times 0.200 \approx 2.087\times10^{-5}\ \text{mol}$. The molar mass is $M_2 = w_2/n_2 = 1.26\ \text{g} / (2.087\times10^{-5}\ \text{ …

Solve $\pi=CRT$ for molar concentration $C$, convert $C$ to moles using the solution volume, then divide …

Do not attempt this molar-mass determination by boiling-point elevation or freezing-point depression instead — for a solute this large, $\Delta T_b$ and $\Delta T_f$ would be far too small to measure …