Colligative Properties
1. Colligative properties depend only on the number - not the kind of solute
particles present
Vapor Pressure
1. The bonds between molecules keep molecules from escaping.
2. In a solution, some of the solvent is busy keeping the solute dissolved.
3. Lowers the vapor pressure. (presence of solute.)
4. Electrolytes form ions when dissolved - more pieces.
5. NaCl Na+ + Cl2 pieces
6. More pieces bigger effect.
Boiling point Elevation
1. The vapor pressure determines the boiling point.
2. Because a non-volatile solute lowers the vapor pressure it raises the
boiling point.
3. Lower vapor pressure - higher boiling point.
4. Salt water boils above 100ºC
Freezing Point Depression
1. Solids form when molecules make an orderly pattern.
2. The solute molecules break up the orderly pattern.
3. Makes the freezing point lower.
4. Salt water freezes below 0ºC
Molality
1. A new unit for concentration
2. m = Moles of solute ÷ kilogram of solvent
3. m = Moles of solute ÷ 1000 g of solvent
How many grams of KI must be dissolved in 500g of water to produce a
0.060 m KI solution?
500 g H2O x (0.60 mol KI /1000 g H2O) = 0.03 mol KI
0.03 mol KI x (166.1 g KI/ mol) = 5.0 g KI
Calculate the molality of a solution prepared by dissolving 10 g NaCl in 600
g of water.
10g NaCl x (1 mol / 58 g NaCl) = 0.1724 mol NaCl
600 g H2O g x (1 kg/1000g) = 0.60 kg H2O
m = 0.1724 mol NaCl/0.60 kg H2O = 0.287 m
Mole Fraction
1. Used to express concentration
Xsolute = moles component / total moles solution
2. The sum of all mole fractions is always equal to one.
What is the mole fraction of SO2 containing 128 g SO2 dissolved in 1500 g
CO2 ?
Convert grams into moles.
Moles SO2 = 1.99 mol
Moles CO2 = 34.09
XSO2 = 1.99 / (1.99 + 34.09) = 0.05515
Electrolytes in solution
1. Since colligative properties only depend on the number of molecules.
2. Ionic compounds should have a bigger effect.
3. When they dissolve they dissociate.
4. Individual Na and Cl ions fall apart.
5. 1 mole of NaCl makes
2 moles of ions.
6. 1 mole Al(NO3)3 makes 4 moles ions.
7. Electrolytes have a bigger impact on melting and freezing points per mole
because they make more pieces.
8. Relationship is expressed using the van’t Hoff factor i
i = (moles of particles in soln) / (moles solute dissolved)
9. If it is a non-electrolyte then i is equal to one
Occurs with covalent compounds
Boiling Point Elevation Formula
T = i Kb m
T = Tf – Ti
i = # particles
m = molality
Kb is a constant which dpends on the solvent
Kb for water = 0.512 °C kg / g
T = i Kf m
Kf = 1.86 °C kg / g
Freezing Point Depression Formula
Examples
1. At what temperature will a solution composed of 0.73 moles of glucose in
650 mL water begin to boil?
Convert 650 mL H2O to kg
Density of H2O is 1g per 1 mL
650 mL H2O = 650 g = 0.650 kg
Calculate the molality: (0.73 mol) / 0.650 kg = 1.1 m
Plug into the formula: ∆T= ikbm
= (1)(0.512 °C kg / g)(1.1 m)
= 0.56 °C
Tells the change in Temp
Boiling Point Elevation add the change in temp to the actual bp of the solvent
100 °C + 0.56 °C = 100. 56 °C
2. At what temperature will a solution of ethylene glycol freeze if it contains
120 g ethylene glycol in 500 mL water?
GFM : 62.1 g/mol
Convert 500 mL H2O to kg = 0.500 kg H2O
Convert g to mol
120g ethylene glycol = 1.93 mol
Calculate the m = 1.93 mol / 0.500 kg = 3.86 m
Formula : T = i Kf m
= (1)(1.86 °C kg / g)(3.86 m)
= 7.17 °C
Normal freezing point for H2O is 0 °C
T = 0 °C – 7.17 °C = - 7.17 °C
3. What mass of NaCl would have to be dissolved in 1000 g water to raise
the boiling point by 2 °C ?
Calculate m using info from the question.
m = T / i Kb
= 2 °C / (2)(0.512 °C kg / g
= 1.95 m
Convert the molality to just moles
1.95 m x 1 kg H2O = 1.95 mol
Use the GFM to calculate grams
1.95 mol x (58 g NaCl/mol) = 114 g
Using Kf or Kb to Find Molar Mass
Molality  moles  molar mass
Usually assume non-electrolyte so i = 1
When 36 g of a substance is dissolved in 100 g H2O, the solution begins to
freeze at -3.72 °C. What is the molar mass?
Find T
- 3.72 °C – 0 = 3.72
Find m
m = 3.72 °C / (1)(1.86 °C kg / g
=2m
Find the mol of solute
2 m x 1 kg H2O = 2 mol
Molar Mass (g/mol):
36 g / 2 mol = 180 g / mol