Question

18 g of glucose `(C_(6)H_(12)O_(6))` is dissolved in 1 kg of water in a saucepn. At what temperature will water boil under 1.013 bar pressure ? Given `K_(b)` for water is 0.52 K kg `mol^(-1)`

Answer 1

`W_(B)=(M_(B)xxDeltaT_(b)xxW_(A))/K_(b) or DeltaT_(b)=(W_(B)xxK_(b))/(M_(B)xxW_(A)`
Mass of solute `(W_(B))`=18g
Mass of water `(W_(B))`=1 kg
Molar mass of solute `(M_(B))`= 0.52 K kg `mol^(-1)`
Molal elevation constant `(K_(b))`=0.52 K kg `mol^(-1)`
`DeltaT_(b)=((18g)xx(0.52 K kg mol^(-1)))/((180 g mol^(-1))xx(1 kg))=0.052 K`
At 1.013 bar pressure (atmospheric pressure), boiling point of water= 37300K
Boiling point of solution= `(373.0+0.052 K)`