Question

18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan.
At what temperature will water boil at 1.013 bar? Kb, for water is 0.52
K kg mol^-1​

Answer 1

Answer:

373.052K

Explanation:

△T=Kb× (WB×1000)/(MB×WA)

△T=0.52× (18×1000)/(180×1000)

T−To=0.052

T−373=0.052

T=373.052K