`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)`.

Answer

`373.202 K`

Explanation

`Delta T_(b) = (k_(b) xx w_(B) xx 100)/(M_(B) xx w_(A))`
` = (0.52 xx 18 xx 1000)/(180 xx 1000) = 0.052`
Boiling point of solution .
`= 373.15+ 0.052 = 373 .202 K`