`18 g` of glucose `(C_(6)H_(12)O_(6))` is dissolved in `1 kg` of water in a saucepan. At what temperature will the water boil (at 1 atm) ? `K_(b)` for water is `0.52 K kg mol^(-1)`.

Explanation

`Delta T_(b) = (K_(b) xx w_(B) xx 1000)/(M_(B) xx w_(A))`
` = (0.52 xx 1.8 xx 1000)/(180 xx 1000)`
`= 0.052^(@)`
Boiling point of solution = 3.73.15+0.052 = 373.202 K.