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200cm^(3) of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of this solution at 300 K is found to be 2.57xx10^(-3) bar. Calculate the molar mass of the protein. |
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Answer» Solution :Here, we are given : Mass of solute (protein), `w_(2)=1.26 g` Volume of the solution (V) `=200cm^(3)=0.200L` `pi=2.57xx10^(-3)" bar, T = 300 K, R = 0.083 L bar K"^(-1)"MOL"^(-1)` Substituting these values in the formula, `M_(2)=(w_(2)RT)/(piV),` we get `M_(2)=(1.26gxx0.083 " L bar K"^(-1)"mol"^(-1)xx300K)/(2.57xx10^(-3)" bar"xx0.200L)="61039 g mol"^(-1)` |
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