Question

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)`.

Answer 1

Moles of glucose `=18"g"//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
`DeltaT_(b)=K_(b)xx m=0.52" K kg mol"^(-1)xx0.1" 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".`