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)
1
Expert's answer
2013-12-20T11:11:59-0500
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
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