18g of glucose, `C_(6)H_(12)O_(6)`, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar ? `K_(b)` for water is 0.52 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
`ÅT_(b) = K_(b) xx m = 0.52 " K 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.