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InterviewSolution
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Dissolution of 1.5 g of a non-volatile solute `(mol.wt.=60)` in 250 g of a solvent reduces its freezing point by `0.01^(@)C`. Find the molal depression constant of the solvent.A. `0.01`B. `0.001`C. `0.0001`D. `0.1` |
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Answer» Correct Answer - D Depression in freezing point, `DeltaT_(f)=K_(f)xxm` `DeltaT_(f)=K_(f)xxm` `"where,m=molality"=("wt.of solute"xx 1000)/("mol.wt.of solute"xxwt.of" solvent")` " `=(1.5xx1000)/(60xx250)=0.1` `DeltaT_(f)=K_(f)xx0.1` `0.01=K_(f)xx0.1` `:.K_(f)=(0.01)/(0.1)=0.1^(@)C kg" mol"^(-1)` |
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