Discussion
100 cm3 of an aqueous solution of a protein contains 1.5 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 4.5 × 10-3 bar. The molar mass of the protein is
[R = 0.083 lit atm mol-1 K-1]
Answer: D
The molar mass is \(8.3 \times 10^{4}\) g mol-1.
Use \(\pi = \frac{n}{V}RT\) with \(n = \frac{w}{M}\), so \(M = \frac{wRT}{\pi V}\). The value 0.083 is the gas constant in L bar mol-1 K-1 (the stem labels it lit atm), which suits the pressure given in bar; V = 0.1 L.
\[ \begin{aligned} M &= \frac{1.5 \times 0.083 \times 300}{4.5 \times 10^{-3} \times 0.1} \\ &= \frac{37.35}{4.5 \times 10^{-4}} \\ &= 8.3 \times 10^{4}\ \text{g mol}^{-1} \end{aligned} \]
\(8.3 \times 10^{5}\) and \(8.3 \times 10^{3}\) come from taking the volume as 0.01 L or 1 L. The two options showing \(4.15 \times 10^{4}\), half the answer, are identical in the paper.
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