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. `0.93^(@)C`
B. `-0.93^(@)C`
C. `1.86^(@)C`
D. `-1.86^(@)C`

Answer

D

Explanation

`Delta T_(b) = 100.512 - 100 = 0.512^(@)`
`m = (Delta T_(b))/(k_(b)) = (0.512)/(0.512) = 1 m `
`Delta T_(f) = k_(f) xx m = 1.86 xx 1 = 1.86^(@)`
Freezing point of solution `= 0 - 1.86 = - 1.86^(@)C` .