A solution of urea in water boills at `100^(@)C`. Calculate the freezing point of the same solution. Molal constant `K_(f) and K_(b)` are 1.86 and 0.521 k `m^(-1)` repectivly.

Answer

`-0.654^(@)C`

Explanation

`DeltaT_(f)=K_(f)xxm" and "DeltaT_(b)xxm`
`(DeltaT_(f))/(DeltaT_(b))=K_(f)/K_(b)`
`DeltaT_(b)=100.18^(@)C-100.0=0.18^(@)C=0.18 K`
`K_(f)=1.86" K m"^(-1), K_(b)=0.512" K m"^(-1)`
`DeltaT_(f)=K_(f)/K_(b)xxDeltaT_(b)=((1.86"Km"^(-1)))/((0.512"Km"^(-1)))xx0.18K=0.654K=0.654^(@)C`
`therefore "Freezing point of soluion"=0^(@)C-0.654^(@)C=-0.654^(@)C.`