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

Question

Dr. Abubacker · Accepted Answer

[Solution from source] π = (w/M)RT/V, so M = wRT/(πV) = 1.26 × 0.083 × 300/(2.57 × 10⁻³ × 0.200) = 31.37/5.14 × 10⁻⁴ ≈ 61038 g mol⁻¹. Forgetting to convert 200 mL to 0.200 L or halving/doubling the volume gives the other values. Answer (2).

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

Explanation

Reviewed by Dr. Abubacker

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