`200 cm^(3)` of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of such a solution at 300 K is found to be `2.57 xx 10^(-3)` bar. Calculate the molar mass of the protein.
(R = 0.083 L bar `"mol"^(-1)K^(-1)`)

Explanation

The given quantities are.
`Pi=2.57 xx 10^(-3)` bar
`V=200 cm^(3)=0.200 ` Litre
`T= 300 K`
`R = 0.083 L " bar mol"^(-1)K^(-1)`
`W_(2) = 1.26 g `
`M_(2) = (W_(2)RT)/(Pi V)`
`=(1.26g xx 0.083 " L bar mol"^(-1)K^(-1) xx 300 K)/(2.57 xx 10^(-3) "bar" xx 0.200" Litre")`
`=61,038.91"g mol"^(-1)`