`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

Here, we are given `w_(2)=18 g, w_(1)=1 kg = 1000 g, K_(b)=0.52 "kg mol"^(-1)`
`M_(2) ("for glucose", C_(6)H_(12)O_(6)) = 72+12+96=180 g mol^(-1)`
`Delta T_(b)=(1000 K_(b)w_(2))/(w_(1)M_(2))=(1000" g kg"^(-1)xx0.52"K kg mol"^(-1)xx18 g)/(1000 g xx 180 " g mol"^(-1))=0.52 K`
As water boils at 37.3 15 K at 1.013 bat pressure, therefore, boiling point of solution
`= 373.15+0.052 K = 373.202 K`.