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 = (wt xx 1000)/(MM xx V) RT`
`2.57 xx 10^(-3) = (1.26 xx1000)/(MM xx 200) xx 0.083 xx 300 `
MM = 6610038 g