`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 1000 xx w_(B))/(w_(A) xx M_(B))`
` w_(B) = 18g , w_(A) = 1000 g , M_(B) = 180`
`k_(b) = 0.52 K m^(-1)`
`therefore " "Delta T_(b) (0.52 xx 1000 xx 18)/(1000 xx 180) = 0.052 K`
Boiling point of solution `= 373 . 15+0.052 = 373.202 K`