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

Answer

D

Explanation

`pi = cRT = [(W_(2))/(M_(2)) xx (1)/(V)] RT` or `M_(2) = (W_(2)RT)/(piV)`
`= (1.26 g xx 0.083 L "bar" mol^(-1)K^(-1) xx 300 K)/(2.57 xx 10^(3)"bar" xx 0.200L)`
`= 61038 g mol^(-1)`.