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A solution of urea in water has a boiling point of 100.128^(@)C. Calculate the freezing point of the same solution. Molal constants for water K_(f) and K_(b) are 1.86^(@)C and 0.512^(@)C respectively. |
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Answer» Solution :Step I. To calculate MOLALITY of the solution from boiling point data : We are GIVEN that`""DeltaT_(b)=100.128-100=0.128^(@)C, K_(b)=0.512^(@)C` ltbRgt Using the formula,`""DeltaT_(b)=K_(b)xxm`, where m is the molality, we get `""m=(DeltaT_(b))/(K_(b))=(0.128)/(0.512)=0.25` Step II. To caluate the DEPRESSION in freezing point : We are given that `K_(F)=1.86^(@)C` m = 0.25 (calculated above) `therefore""DeltaT_(f)=K_(f)xxm=1.86xx0.25=0.465^(@)` `therefore"Freezing point of the solution "(T_(f))=T_(f)^(@)-DeltaT_(f)=0^(@)C-0.465^(@)=-0.65^(@)C` |
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