Freezing point of an aqueous solution is `-0.186^(@)C`. Elevation of boiling point of the same solution is ……..if `K_(b)=0.512 K "molality"^(-1)`and `K_(f)=1.86K "molality"^(-1)` :

Options

A. `0.186^(@)C`
B. `0.0512^(@)C`
C. `0.092^(@)C`
D. `0.237^(@)C`

Answer

B

Explanation

`cancelOT_(f)=(1000.K_(f).w)/(W.m)`and`DeltaT_(b)=(1000.K_(b).w)/(W.m)`
`:.(cancelOT_(f))/(cancelOT_(b))=(K_(f))/(K_(b))` or `(0.186)/((cancelOT_(b)))=(1.86)/(0.512)`
`DeltaT_(b)=0.0512`.