Question #37595

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)

Expert's answer

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

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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