When `0.6g` of urea dissolved in `100g` of water, the water will boil at `(K_(b)` for water `= 0.52kJ. mol^(-1)` and normal boiling point of water `=100^(@)C)`:

Options

A. 372.48 K
B. 273.52 K
C. 373.052 K
D. 273.052 K.

Answer

C

Explanation

`" "Delta T_(b) = (k_(b) xx 1000 xx w_(B))/(w_(A) xx M_(B))`
` = (0.52 xx 1000 xx 0.6)/(100xx 60) - 0.052`
`therefore` Boiling point of solution `: 373+0.052=373.052 K`