100 mL of an aqueous solution of protein contais 6.3 g of protein. Calculate the molar mass of protein if osmatic pressure of the solution at `27^@C` is `2.57 xx 10^(-3)` bar.

Explanation

Osmotic pressure `pi=2.57xx10^(-3)` bar ,
weight of solute = w= 6.3 g
Solution constant =S=0.083 L bar `K^(-1) "mol"^(-1)`
Volume of solution =V=100mL = 0.1 L : Absolute temperature =300 K
Molar mass =`(wST)/(piV)=(6.3xx0.083xx300)/(2.57xx10^(-3)xx0.1)="610 g mol"^(-1)`