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200 mL of an aqueous solution of a protein contains its 1.26 g. 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).

1. 51022 g mol⁻¹
2. 122044 g mol⁻¹
3. 31011 g mol⁻¹
4. 61038 g mol⁻¹

Correct answer: 122044 g mol⁻¹

Solution

The osmotic pressure equation is π = CRT. Rearranging for molar mass (M), M = (wRT) / (πV). Substituting values: w = 1.26 g, R = 0.083 L bar mol⁻¹ K⁻¹, T = 300 K, π = 2.57 × 10⁻³ bar, and V = 0.2 L, we get M = (1.26 × 0.083 × 300) / (2.57 × 10⁻³ × 0.2) = 122044 g mol⁻¹.

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