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 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

We know that, `piV = nRT, " where " n =w/M`
= `piV = w/M RT`
`M = (wRT)/(piV) = (1.26xx0.083 xx 300)/(2.57xx10^(-3)xx 200/1000)`
`=(1.26 xx0.083 xx 300)/(2.57xx10^(-3)xx0.2)=61038 g mol^(-1)`