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300cm^(3) of an aqueous solution of a pr...

`300cm^(3)` of an aqueous solution of a protein contains 2.12 g of the protein, the protein, osmotic pressure of such a solution at 300 K is found to be `3.89xx10^(-3)" bar."` Calculate the molar mass of the protein. `("R = 0.0823 L bar mol"^(-1)K^(-1))`

Text Solution

Verified by Experts

`M_B = (W_BRT)/(piV)`
`= (2.12 xx 0.0823 xx 300)/(3.89 xx 10^(-3)xx0.3)`
`= 44852.4`
Detailed Answer :
Give `V =300 cm^3`
`= 0.300 "litre"`
T =300 K`,
`pi = 3.89 xx 10^(-3) "bar"`
`R = 0.0723 L "bar" mol^(-1) K^(-1)`,
`W_2 =2 .12g`
`M_2 =(WRT)/(piV)`
`M_2 = (2.12 xx 0.0823 xx 300)/(3.89 xx 10^(-3) xx 0.3)`
`M_2 = 44852.4 g//mol`.
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