1.

Calculate the freezing point depression and boiling point elevation of a solution of `10.0 g` of urea `(M_(B)=60)` in `50.0 g` of water at `1 atm`. pressure.`K_(b)` and `K_(f)` for water `0.52^(@)C m^(-1)` and `1.86^(@)C m^(-1)` respectively.

Answer» Depression in freezing point =`DeltaT_(f)` and elevation in boiling point =`DeltaT_(b)`
`DeltaT_(f)=K_(f)((W_(B)/(Mw_(B)))/W_(A)xx1000)=1.86(((10)/(60))/(50)xx1000)=6.2^(@)C`
`DeltaT_(b)=K_(b)((W_(B)/(Mw_(B)))/W_(A)xx1000)=0.52(((10)/(60))/(50)xx1000)=1.7^(@)C`
Freezing point `(T_(f))=T_(f)^(@)-DeltaT_(f) =0-6.2 =-6.2^(@)C`
Boiling point `(T_(b))=T_(b)^(@)-DeltaT_(b) =100 + 1.7 =-101.7^(@)C`


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