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

Options

A. `31011 g mol^(-1)`
B. `61038 g mol^(-1)`
C. `51022 g mol^(-1)`
D. `122044 g mol^(-1)`

Answer

B

Explanation

`pi=CRT=(wtxx1000)/(GMMxxV)RT`
`2.57xx10^(-3)=(1.26xx1000)/(GMMxx200)xx0.083xx300`
GMM = 61038 g