`200" cm"^(2)` of an aqueous solution of a protein contains 1.26 g of the protein. The oxmotic 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:
` ð = 2.57 xx10^(-3) ` bar .
V = 200 `cm^(3) =0.200 ` litre
T = 300 K
R = 0.083 L bar `mol^(-1) K^(-1)`
Substituting these values in equation (2.42) we get
`M_(2) = (1.26 g xx 0.083 " L bar K"^(-1) mol^(-1) xx300 K )/( 2.57 xx10^(-3) "bar" xx 0.200 L ) = 61. 022 " g mol"^(-1)`