18 g glucose, `C_(6)H_(12)O_(6)`, is dissolved in 1 kg of water in saucepan. At what temperasture will water boil at 1.013 bar ? `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 = 1 kg
Thus molality of glucose solution `= 0.1 mol kg^(-1)`
For water, change in boiling point
`Delta T_(b)=K_(b)xx m`
`=0.52 "K kg mol"^(-1)xx 0.1 mol kg^(-1) = 0.052` K
Since water boils at 373.15 K at 1.013 bar pressure. Therefore, the boiling point of solution will be `373.15+0.052=373.202 K`.