1.

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 300K is found to be 2.57xx10^(-3) bar. Calculate the molar mass of the protein.

Answer»

Solution :The VARIOUS quantities known to us are as follows :
`pi = 2.57xx10^(-3)` bar,
`V=200 CM^(3)=0.200` litre
T = 300 K
`R = 0.083 " L bar mol"^(-1)K^(-1)`
Putting above value in equation :
`M_(2)=(w_(2)RT)/(pi V)`
`M_(2)=(1.26 g XX 0.083 " L bar K"^(-1)mol^(-1)xx 300K)/(2.57xx10^(-3)"bar"xx0.200 L)`
`= 61,022 g mol^(-1)`


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