Question

An aqueous solution of volume 300 cm³ 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

Answer 1

Correct Answer: 40535

Given: Volume (V) = 300 cm³ = 0.3 L Mass (m) = 0.63 g Osmotic pressure ($\pi$) = 1.29 mbar = 1.29 × 10⁻³ bar Temperature (T) = 300 K R = 0.083 L bar K⁻¹ mol⁻¹ Using the formula: $\pi = cRT$, where c is the molarity. Calculate molarity (c): $$c = \frac{\pi}{RT} = \frac{1.29 \times 10^{-3} \text{ bar}}{0.083 \text{ L bar K}^{-1} \text{mol}^{-1} \times 300 \text{ K}}$$ $$c \approx 5.18 \times 10^{-5} \text{ mol/L}$$ Calculate moles (n): $$n = c \times V = 5.18 \times 10^{-5} \text{ mol/L} \times 0.3 \text{ L}$$ $$n \approx 1.554 \times 10^{-5} \text{ mol}$$ Calculate molar mass (M): $$M = \frac{m}{n} = \frac{0.63 \text{ g}}{1.554 \times 10^{-5} \text{ mol}}$$ $$M \approx 40540 \text{ g/mol}$$ The molar mass of the protein is approximately 40540 g/mol.