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)
- 51022 g mol-1
- 122044 g mol-1
- 31011 g mol-1
- 61038 g mol-1
Answer
(4) 61038 g mol-1
Solution
V = wMRT ⇒ M = wRT V
M = 1.26 0.083 3002.57 10-3 0.200
= 31.3745.14 10-4
61038 g mol-1
Answer: (4)