A solution of urea (mol. Mass `60 g mol^(-1)`) boils of `100.18^(@)C` at one one atmospheric pressure. If `k_(f)` and `K_(b)` for water are `1.86` and `0.512 K kg mol^(-1)` respectively, the above solution will freeze at:

Options

A. `-6.54^(@)C`
B. `6.54^(@)C`
C. `0.654^(@)C`
D. `-0.654^(@)C`

Answer

4

Explanation

We have
`DeltaT_(b)=iK_(b)m`
`DeltaT_(f)=iK_(f)m`
Taking the ratio, we get
`(DeltaT_(b))/(DeltaT_(f))=K_(b)/K_(f)`
`(0.18^(@)C)/(DeltaT_(f))=(0.512^(@)Cm^(-1))/(1.86^(@)Cm^(-1))`
`(DeltaT_(b)=T_(b)-T_(b)^(0)=100.18^(@)C-100.00^(@)C=0.18^(@)C)`
or `DeltaT_(f)=0.6539^(@)C=0.654^(@)C`
Freezing point of solution=Freezing point of water-`DeltaT_(f)`
`=0.000^(@)C-0.654^(@)C`
`=-0.654^(@)C`