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. 0.186
B. 0.512
C. 0.512//1.86
D. 0.0512

Answer

D

Explanation

`Delta T_(b) = K_(b) xx "molality"`
`Delta T_(b) = K_(b) xx"molality" `
`(Delta T_(b))/(Delta T_(f))=K_(b)/K_(f) ` or `Delta T_(b) = (Delta T_(f) xx K_(b))/K_(f)`
`=(0.186 xx 0.512)/1.86=0.0512`