200 mL of an aqueous solution of a protein contain 1.26 g of protein, the osmotic pressure of this solution at 300 K is found to be 2.57 10⁻³ bar, Calculate the molar mass of protein. ( R = 0.083 bar mol ⁻¹ K ⁻¹ )

Step-by-step solution

Given: Volume of solution, V = 200 mL = 0.200 L Mass of protein, W = 1.26 g Osmotic pressure, = 2.57 10⁻³ bar Temperature, T = 300 K Gas constant, R = 0.083 L bar K ⁻¹ mol ⁻¹ Using the formula for osmotic pressure: = W M V R T Rearranging for molar mass M : M = W R T V Substituting the values: M = 1.26 0.083 300 2.57 10⁻³ 0.200 M = 31.374 5.14 10⁻⁴ M = 61038.9 g mol ⁻¹ 61039 g mol ⁻¹ Answer: 61039 g mol ⁻¹