`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 `2.57xx10^(-3)` bar. Calculate the molar mass of the protein.

Explanation

The various quantities known to us are as follows :
`pi = 2.57xx10^(-3)` bar,
`V=200 cm^(3)=0.200` litre
T = 300 K
`R = 0.083 " L bar mol"^(-1)K^(-1)`
Putting above value in equation :
`M_(2)=(w_(2)RT)/(pi V)`
`M_(2)=(1.26 g xx 0.083 " L bar K"^(-1)mol^(-1)xx 300K)/(2.57xx10^(-3)"bar"xx0.200 L)`
`= 61,022 g mol^(-1)`