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1.0 g protein in 100 mL gives 2.49 × 10^-3 bar at 300 K (R 0.083). Molar mass?
A10,000 g/mol
B1,000,000 g/mol
C100,000 g/mol
D1,000 g/mol
Answer & Solution
Correct answer: C. 100,000 g/mol
M = wRT/(pi V) = (1.0 × 0.083 × 300)/(2.49 × 10^-3 × 0.100) = 1.0 × 10^5 g/mol.
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