`300cm^(3)` of an aqueous solution of a protein contains 2.12 g of the protein, the protein, osmotic pressure of such a solution at 300 K is found to be `3.89xx10^(-3)" bar."` Calculate the molar mass of the protein. `("R = 0.0823 L bar mol"^(-1)K^(-1))`

Explanation

`M_(2)=(W_(2)RT)/(piV)=(2.12xx0.0823xx300)/(3.89xx10^(-3)xx0.3)" "["300 cm"^(3)=0.32]`
`M_(2)=44.852xx10^(3)=44852g//ml`