An aqueous solution of volume 300 cm$$^3$$ contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is _______ g mol$$^{-1}$$.
Given: R = 0.083 L bar K$$^{-1}$$ mol$$^{-1}$$

We need to find the molar mass of a protein given its osmotic pressure data.

Volume = 300 cm$$^3$$ = 0.3 L, mass of protein = 0.63 g, Temperature = 300 K, Osmotic pressure $$\pi = 1.29$$ mbar = $$1.29 \times 10^{-3}$$ bar, $$R = 0.083$$ L bar K$$^{-1}$$ mol$$^{-1}$$.

The van't Hoff equation for osmotic pressure is:

$$\pi = CRT = \frac{n}{V}RT = \frac{m}{MV}RT$$

where $$C$$ is molar concentration, $$n$$ is moles of solute, $$m$$ is mass of solute, $$M$$ is molar mass, $$V$$ is volume in liters.

$$M = \frac{mRT}{\pi V}$$

$$M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.3}$$

$$0.63 \times 0.083 = 0.05229$$

$$0.05229 \times 300 = 15.687$$

$$1.29 \times 10^{-3} \times 0.3 = 3.87 \times 10^{-4}$$

Divide.

$$M = \frac{15.687}{3.87 \times 10^{-4}} = \frac{15.687}{0.000387} \approx 40535 \text{ g/mol}$$

The molar mass of the protein is 40535 g mol$$^{-1}$$.