18 g of glucose, C6H12O6 (Molar mass = 180 g mol-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water =0.52 K kg mol-1, boiling point of pure water =373.15K)
Weight of solvent H2O (W1) = 1kg = 1000gm
Molar mass of solvent H2O = 18gm
Weight of solute (C6H12O6 ) (W2) = 18gm
Molar mass of solute (M2) = 180gm
∴ Moles of solute =
= 0.1moles
Kb for water =0.52 K kg mol-1
Boiling point of pure water =373.15K
∴ Elevation in boiling point ΔTb = 
= ![]()
=0.052K
∴ ΔTb = Tb – T°b [Tb = Boiling point of solution]
[T°b =Boiling point of pure water]
0.052 = Tb – 373.15
∴ Tb = 373.202K
Hence, boiling point of the solution is 373.202K
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