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

Explanation

The various quantities known to us are as follows `prod=2.57xx10^(-3)` bar,
`V=200cm^(3)=0.200` litre
`T=300K`
`R=0.083L` bar `mol^(-1)" "K^(-1)`
Substituting these values in equation we get
`M_(2)=(1.26gxx0.083" L bar "K^(-1)" "mol^(-1)xx300K)/(2.57xx10^(-3)"bar"xx0.200L)=61.022g" "mol^(-1)`