1.

A `10 m` solution of urea is cooled to `-13.02^(@)C`. What amount of urea will separate out if the mass of solution taken is `100 g` ? `[K_(f)("water") =1.86 Kkg mol^(-1)]`

Answer» `DeltaT_(f)=0-(-13.02)=13.02 K`
`m=(DeltaT_(f))/(K_(f))`
`:.m=(13.02)/(1.86)=(7.0 mol kg^(-1))`
`:. m=10=(W_(2)xx1000)/(60xx(100-W_(2)))`
`:. W_(2)=37.5 g`
Mass of `H_(2)O` in this solution =`(100-37.5)=62.5 g`
The mass of water does not change as urea separates out to bring down to molality to `7 mol kg ^(-1)`.
`:. m=(W_(2)xx1000)/(Mw_(2)xxW_(1))`
`:. 7=(W_(2)xx1000)/(60xx62.5) rArr W_(2)=26.25 g`
Mass of urea separated =`(37.5-26.25)=11.25g`


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