`1000gm H_(2)O` have `0.1` mole urea and its freezing point is-`0.2^(@)C` and low now it is freezed upto`-2^(@)C` then how much amount of ice will form.

Explanation

It is assume that solute do not freeze and do not vapurise
`DeltaT_(F)=0.2=K_(t)(0.1)/(1000)xx1000` .....(i)
`DeltaT_(F)=2=K_(t)(0.1)/("wt. of solvent)xx1000` .....(ii) on dividing `("wt. of solvent")/(1000)=(0.2)/(2)`
weight of remaining `H_(2)O` is `100gm` and weight of ice is `900gm`.