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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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