For an aqueous solution freezing point is `-0.186^(@)C`. The boiling point of the same solution is `(K_(f) = 1.86^(@)mol^(-1)kg)` and `(K_(b) = 0.512 mol^(-1) kg)`

Options

A. `0.186^(@)`
B. `100.0512^(@)`
C. `1.86^(@)`
D. `5.12^(@)`

Answer

B

Explanation

`(DeltaT_(b))/(DeltaT_(f)) = (K_(b))/(K_(f)) rArr DeltaT_(b) = (0.512)/(1.86) xx 0.186 = 0.0512^(@)C`
`T'_(b) = T_(b) +DeltaT_(b) = 100.0.512^(@)C`