Molarity and Colligative
Properties
Chemistry GT 5/8/15
Drill
• Use the table from pg. 9 to give the
amount of substances that will dissolve in
100 g of H2O:
– NH3 at 10°C and 80°C
– Why does solubility of NH3 decrease at higher
temperatures?
– KCl at 10°C and 70°C
– If 50 g of KCl were dissolved in 100 g of
water, and it was cooled to 50°C, what kind
of solution would you have?
• HW:
– pg. 5 Molality Problems
– 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
• Drill
• Molality Notes & Example
• Colligative Properties Research
Chart
• Colligative Properties Notes
• Colligative Properties Calculations
• Exit Ticket
Molality
• Don’t you mean molarity??
• Nope! Molality is another way to
measure concentration.
• Equation: m = n
m
• n = moles of solute
• m = mass of solvent in kilograms
Why??
• Molarity changes with the density of
the solvent, so at different
temperatures, you have different
molarities.
• Molality does NOT change with
density or temperature!
• Let’s do #1 on pg. 5
What’s next?
COLLIGATIVE
PROPERTIES!
Colligative Properties
Chart
• You have 15 minutes to fill in the
chart of notes.
• Split the work up at your table—each
person can take one quadrant, and
then share.
• You may use a textbook or BYOD
• We will review the chart together
Colligative Properties
Notes
Colligative Properties
•
A property that depends on the
number of molecules present, but
not on their chemical nature
There are 4 colligative
properties
•
•
•
•
Vapor Pressure
Boiling Point
Freezing Point
Osmotic Pressure
Vapor Pressure
•
•
Liquid molecules at the surface of
a liquid can escape to the gas
phase
- This process is reversible (g  l)
Vapor pressure of a solution
containing a nonvolatile solute is
less than the pressure of the
solvent alone
Vapor Pressure
Solute particles take up
surface area and lower the
vapor pressure
Vapor Pressure
• Vapor pressure reduction is
proportional to the
concentration of the solution
• When the concentration goes
up, the vapor pressure is
reduced
Vapor Pressure
• This partially explains why The
Great Salt Lake has a lower
evaporation rate than expected.
The salt concentration is so high
that the vapor pressure (and
evaporation) has been lowered
Freezing Point
•
The freezing point of a solution is
always lower than that of the
solvent alone
This is called Freezing Point
Depression
•
–
Explains why salt (CaCl2) is used on
ice
- The salt solution that forms has
a lower freezing point than the
original ice
Boiling Point
• The boiling point of a solution
is always higher than that of
the solvent alone
• Boiling Point Elevation
–
You continue to use antifreeze in the
summer, because you want the
coolant to boil at a higher
temperature so it will absorb the
engine heat
Osmotic Pressure
• The tendency for a solution to take
in water due to osmosis
– This is why cells have to maintain their
internal pressures—if they stop
pumping water out, they will fill up and
pop!
Boiling and Freezing
Point Calculations
BEFORE WE CALCULATE…
WE NEED TO TALK ABOUT
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°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°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°C/m x .239 m = .444°C
• Ba(NO3)2  Ba+2 + 2 NO3-1 = 3 moles of ions (i
value)
• .444°C x 3 = 1.33°C
• 0°C – 1.33°C = -1.33°C
Exit Ticket
• Determine the Van’t Hoff Factor for
the following compounds.
– AlCl3
– Mg3(PO4)2
– C6H12O6