Answer to Question #37595 in Physical Chemistry for saurabh

Question

18 g glucose, C6 Hl2O6 (Molar mass = 180 g mol) is dissolved in 1 kg of water in a sauce pan. At
temperature will this solution boil (kb for water = 0.52K kg mol
-1
B.P of pure water = 373.1 K)

Answer 1

ΔT = μ*Kb,
where μ is the molality:
μ = n(solute)/m(solvent)
n = m/M = 18/180 = 0.1 mol
μ = 0.1/1 = 0.1 mol/kg

ΔT = 0.1*0.52 = 0.052; hence Tb = 373.10 + 0.05 = 373.15 K
Tb = 373.15 K