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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. |
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Answer» The given values are `pi=2.57xx10^(-3)` bar `V=200 cm^(3)=0.2 L` `T=300 K` `R=0.083 L` bar `mol^(-1) K^(-1)` Using the formula, `Mw_("solute")=(W_("solute")xxRxxT)/(pixxV)` `= (1.26 g xx 0.083 L` "bar" `"mol"^(-1) K^(-1)xx300 K)/(2.57xx10^(-3) "bar"xx0.2 L)` `= 61022 g mol^(-1)` Thus, the molecular weight of solute is `61022 g mol^(-1)`. |
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