`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

Moles of glucose `=18g//180"g "mol^(-1)=0.1` mol
Number of kilograms of solvent `=1kg`
Thus molality of glucose solution `=0.1` mol `kg^(-1)`
For water, change in boiling point.
`DeltaT_(b)=K_(b)xxm=0.52"K kg "mol^(-1)xx0.1" mol "kg^(-1)=0.052K`
Since water oils at 373.15K at 1.013 bar pressure, therefore, the boiling point of solution will be `373.15+0.052=373.202K`