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Chemistry lesson about freezing point and boiling point with exercises
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COLLIGATIVE PROPERTIES:FREEZING POINT DEPRESSION AND BOILING POINT ELEVATIONDAY 1 – 04 FEBRUARY 2015
Colligative Properties
Depends on the NUMBER of solute, not on the nature of solute particles
Freezing Point Depression Boiling Point ElevationVapor Pressure LoweringOsmotic Pressure
Electrolyte and Non- electrolytes Electrolytes
•Separates in water forming a solution that conducts electric current
•IONIC COMPOUNDS
Non- electrolytes
•does not allow the flow of an electric current
•COVALENT COMPOUNDS
Freezing Point Depression
Freezing Point Depression
Adding a solute to a solvent decreases freezing point of the solvent
Tf° (pure solvent) > Tf (solution) Freezing point of pure solvent – 0o
Tf = Tf° - Tf
Amount of solute
Freezing Point of solvent
Freezing Point Depression
Temperature
Freezing Point Depression constant (1.860 C /m)
Molality (mol/kg)
Formula (non-electrolytes):
Freezing Point Depression
ΔTf freezing point temperature
i van ‘t Hoff factor (sum of subscripts)
Kf Freezing Point Depression constant (1.860 C /m)
m Molality (mol/kg)
ΔTf = i Kf m
Formula (electrolytes):
Freezing Point DepressionEXAMPLES
ΔTf freezing point temperature
i van ‘t Hoff factor (sum of subscripts)
Kf Freezing Point Depression constant (1.860 C /m)
m Molality (mol/kg)
NaCl
Na Cl 1 + 1
i = 2
CaCl2
Ca Cl2
1 + 2
i = 3
Na3PO4
Na3 (PO4) 3 + 1
i = 4
PRACTICE
What is the new freezing point of 200 g of water (Kf = -1.86 oC) if195 g of
sucrose (C12H22O11) are added to it?
ANSWER
∆Tf = -5.301°C
PRACTICE
What is the molality of barium sulfate (Ba2SO4) with a freezing a point of
1.12°C?
ANSWER
Molality = 0.20 m
PRACTICE
A solution of 3.39 g of an unknown compound in 10.00 g of water has a freezing point of 7.31°C. The solution does not conduct electricity. What
is the molar mass of the compound?
ANSWER
Molar mass = 86.26 g/mol
Boiling Point Elevation
Boiling Point Elevation Temperature difference between a solution’s boiling point and a pure solvent’s boiling point
For nonelectrolytes:
Tb (solution) > Tb° (pure solvent) Boiling Point of pure solvent - 100°C
Tb = Tb° + Tb (+)
Amount of solute
Boiling Point Elevation
Boiling Point Elevation
Boiling point of elevation
Molal boiling elevation constant (0.5120 C /m)
Molality (mol/kg)
Formula (non-electrolyte):
Boiling Point Elevation
ΔTb freezing point temperature
i van ‘t Hoff factor (sum of subscripts)
Kb Molal Boiling Point Elevation constant (0.5120 C /m)
m Molality (mol/kg)
ΔTb = i Kb m
Formula (electrolytes):
PRACTICE
What will be the boiling point of an aqueous solution containing 55.0 g of
glycerol, C3H5(OH)3, and 250 g of water? Kb(H2O)= 0.512 °c/m
ANSWER
∆Tb = 1.23°C
PRACTICE
What is the boiling point of a solution containing 34.3 g of magnesium nitrate
dissolved in 0.107 kg of water?
ANSWER
∆Tb = 3.3°C
PRACTICE
A solution containing 28.4 g of sodium bromate dissolved in water. Find the
mass of the solvent if the boiling point of the solution is 8.5°C.
ANSWER
Mass of solvent = 0.034 kg
