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`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.A. `61038 g mol^(-1)`B. `51022 g mol^(-1)`C. `122044 g mol^(-1)`D. `31011 g mol^(-1)` |
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Answer» Correct Answer - 1 we have `pi=CRT=n_(2)/VRT=(W_(2)RT)/(M_(2)V)` or `M_(2)=(W_(2)RT)/(piV)` Substituting the given values, we get `M_(2)=((1.26 g)(0.083 L "bar" mol^(-1) K^(-1))(300 K))/((2.57xx10^(-3))(200xx10^(-3)L))` `=61038 g mol^(-1)` |
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