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
