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

Answer

`61022" g mol"^(-1)`

Explanation

`pi=2.57 xx 10^(-3)" bar ", V=200cm^(3)=0.200"litre", T=300K,`
R = 0.083 L bar `"mol"^(-1)K^(-1)`, Mass of protein = 1.26 g
`pi=(W)/(M_(2)) xx (RT)/(V) or M_(2)=(W)/(pi) xx (RT)/(V)`
Substituting the values,
`M_(2)=(1.26g xx 0.083"L bar K"^(-1)"mol"^(-1) xx 300K)/(2.57 xx 10^(-3)"bar" xx 0.200L)`
`=61,022"g mol"^(-1)`