Question

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)`:
A. `373.052 K`
B. `273.52K`
C. `372.48K`
D. `273.052K`

Answer 1

Correct Answer - A
`DeltaT_(b) = ((0.6)/(60))/(100) xx 1000 xx 0.52`
`= (1)/(1000) xx 1000 xx 0.52 = 0.52`
`T - 373 = 0.52 T = 373.52`