An aqueous solution freezes at `-0.186^(@)C (K_(f)=1.86^(@)` ,`K_(b)=0.512^(@)`. What is the elevation in boiling point?

Options

A. `0.0512^(@)C`
B. `100.0512^(@)C`
C. `-0.0512^(@)C`
D. None of these

Answer

A

Explanation

Elevation in boilling point, `DeltaT_(b)=K_(b)xxm`
Depression in freezing point
`DeltaT_(f)=K_(f)xxm`
`(DeltaT_(b))/(DeltaT_(f))=(K_(b))/(K_(f))and(DeltaT_(b))/(0.186)=(0.512)/(186)`
`DeltaT_(b)=0.0512^(@)C`