5/11 Colligative Properties

```Colligative Properties
Chemistry GT 5/11/15
Drill
• List the four colligative properties.
Give a real world example of each.
• What is the molality of a solution
made with 25.0 g LiBr and 750. g of
water?
• HW: Effect of a Solute on FP and BP
Objectives
• Today I will be able to:
– Calculate the molality of a solution
– Describe the 4 colligative properties of
vapor pressure, boiling point, freezing
point and osmotic pressure
– Calculate the Van’t Hoff Factor for a
Compound
– Calculate the freezing point depression
and boiling point elevation of a solute
Agenda
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Drill
Finish Colligative Properties Notes
Colligative Properties Calculations
Exit Ticket
Boiling and Freezing
Point Calculations
BEFORE WE CALCULATE…
THE VAN’T HOFF FACTOR
Van’t Hoff Factor
• Determines the moles of particles that are
present when a compound dissolves in a
solution
• Covalent compounds do not dissociate
– C12H22O11
• 1 mole (Van’t Hoff Factor = 1, the same for all
nonelectrolytes)
• Ionic Compounds can dissociate
– NaCl  Na+ + Cl• 2 moles of ions (Van’t Hoff Factor = 2)
– CaCl2  Ca+2 + 2 Cl• 3 moles of ions (Van’t Hoff Factor = 3)
Determine the Van’t Hoff
Factor for the following
Compounds
• C6H12O6
• KCl
• Al2O3
• P2O5
Calculating Boiling and
Freezing Points
 Tb = Kb  m i
• Kb is the molal boiling point elevation
constant, which is a property of the
solvent
– Kb (H2O) = 0.52&deg;C/m
• m is molality
• i is the Van’t Hoff Factor
• Tb is added to the normal boiling point
Calculating Boiling and
Freezing Points
 Tf = Kf  m  i
• Kf is the molal freezing point
depression constant, which is a
property of the solvent
– Kf (H2O) = 1.86&deg;C/m
• m is molality
• i is the Van’t Hoff Factor
• Tf is subtracted from the normal
freezing point
Boiling and Freezing
Points and Electrolytes
• What is the expected change in the freezing
point of water in a solution of 62.5 grams of
barium nitrate, Ba(NO3)2, in 1.00 kg of water?
• ∆Tf = Kf  m  i
• 62.5 g Ba(NO3)2  .239 moles
• .239 moles/1.00 kg = .239 m
• 1.86&deg;C/m x .239 m = .444&deg;C
• Ba(NO3)2  Ba+2 + 2 NO3-1 = 3 moles of ions (i
value)
• .444&deg;C x 3 = 1.33&deg;C
• 0&deg;C – 1.33&deg;C = -1.33&deg;C
Example problems
• Take out the Freezing Point and
Boiling Point practice WS
• Work the odd # problems on your
own paper—check as you go
Exit Ticket
• Determine the Van’t Hoff Factor for
the following compounds.
– AlCl3
– Mg3(PO4)2
– C6H12O6
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