`200 cm^(3)` of an aqueous solution of a protein contains `1.26 g `of the protein. The osmotic pressure of such a solution at `300 K` is found to be `2.57 xx10^(-3)` bar. Calculate the molar mass of the protein.

Options

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

Answer

1

Explanation

we have
`pi=CRT=n_(2)/VRT=(W_(2)RT)/(M_(2)V)`
or `M_(2)=(W_(2)RT)/(piV)`
Substituting the given values, we get
`M_(2)=((1.26 g)(0.083 L "bar" mol^(-1) K^(-1))(300 K))/((2.57xx10^(-3))(200xx10^(-3)L))`
`=61038 g mol^(-1)`