An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:

Options

A. `0.0585 K`
B. `0.0512 K`
C. `1.864 K`
D. `0.0265K`

Answer

B

Explanation

`0.186 = 1.86 xm`
`m = 0.1`
`DeltaT_(b) = 0.1 xx 0.512 = 0.0512`