200 `cm^3` of an aqueous solution of a protein contains 1.26g of the protein . The osmotic pressure of such a solution at 300K is found to be `2.7 xx 10^-3` bar. Calculate the molar mass of the protein (R=0.083 L bar `mol^-1 K^-1`)

Explanation

`V = 200 cm^2 = 0.2 L , W_B = 1.26 g , pi = 2.7 xx10^(-3) ` bar `T = 300 K`
`pi = CRT ` [Osmotic Pressure equation]
`pi = n/V(RT)`
`implies n = (pi xxV)/(RT)=((2.7xx10^(-3))(0.2))/((0.083)(300))`
`=((2.7xx10^(-3))(0.2))/((0.083)(300))=2.169xx10^(-4)" mol"`
`:.` Number of mole `n=W_B/M_B`
`M_B = W_B/n=(1.26)/(2.169xx10^(-4))=5.809xx10^5 "g mol"^(-1)`
`M_B = 580900 " g mol"^(-1)`