`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.

Explanation

Here, we are given : Mass of solute (protein), `w_(2)=1.26 g`
Volume of the solution (V) `=200cm^(3)=0.200L`
`pi=2.57xx10^(-3)" bar, T = 300 K, R = 0.083 L bar K"^(-1)"mol"^(-1)`
Substituting these values in the formula, `M_(2)=(w_(2)RT)/(piV),` we get
`M_(2)=(1.26gxx0.083 " L bar K"^(-1)"mol"^(-1)xx300K)/(2.57xx10^(-3)" bar"xx0.200L)="61039 g mol"^(-1)`