What is the boiling point of a solution of 1.20g of glycerol, C3H8O3, in 20.0 g of water? What is the freezing point?

TB solution = 100.3315 °C

T freezing solution = -1.209 °C

Given

1.2 g glycerol

20 g water

Required

the boiling point

the freezing point

Solution

ΔTB = KB . m

ΔTF = KF . m

mass water = 20 g = 0.02 kg

Kb water = 0.51 °C/m

KF water = 1.86 °C/m

mol glycerol :

= 1.2 g : 92,09382 g/mol

= 0.013

molal  : the number of moles of solute in 1 kg of solvent  

molal of solution :

= 0.013 : 0.02

= 0.65 m

ΔTB = KB . m

ΔTB = TB solution - TB solvent  

TB solution - 100 = 0.51 x 0.65

TB solution - 100 = 0.3315

TB solution = 100.3315 °C

ΔTF = KF . m

ΔT = KF . m

ΔTF = 1.86 x 0.65

ΔTF = 1.209

T_f  = -1.209 °C

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