An aqueous solution boils at `100.50^(@)C`.The freezing point of the solution would be `(K_(b)` for water `=0.51^(@)C//m),(K_(f)` for water `=1.86^(@)C//m`) [no association or dissociation]

Options

A. `0^(@)C`
B. `-1.86^(@)C`
C. `-1.82^(@)C`
D. `+1.82^(@)C`

Answer

B

Explanation

`DeltaT_(b)=ixxmxxK_(b)rArr0.5=Ixx0.51rArrixxm=0.5//0.51`
`0.5=ixxmxxK_(f)=(0.5)/(0.51)xx1.86=1.82`
`:.F.P.=0-182=-1.82^(@)C`
where m is the solution and`K_(b)`ismolal elevation constant.