Dilution Review
• Take 5 minutes with your partner to
solve the scenario below.
• Oops! I spilled 100mL of water into my
300mL of 6 M HCl. Is my solution too
weak to use if my lab requires a 1M to
1.5 M HCl solution?
• How do you know?
Colligative Properties
Defining Colligative Properties
• PHYSICAL properties of the solvent
that change when a solute is added
– The degree of change is dependent upon
the number of solute particles added to
the solvent
– The more solute particles = the greater the
change in colligative property
Number of Particles of Solute
• Think back to what you’ve learned
about ionic and covalent compounds
dissolving in water.
– Ionic compounds DISSOCIATE as they
dissolve.
– Covalent compounds do NOT dissociate
as they dissolve.
• Apply this idea.
– How many solute particles will 1 piece of
NaCl add to a solvent?
• Answer: 2 (Na+ and Cl-)
Applying the Concept Further
• How many solute particles will 1 piece of
magnesium phosphate add to a solvent?
– Mg3(PO4)2 breaks into 3 Mg2+ ions and 2 PO43ions. Answer: 5 solute particles
• How many solute particles will 1 piece of
sugar add to a solvent?
– Sugar is covalent. Answer: 1 solute particle
• Finally, which of the two above will affect a
colligative property the most?
Two Colligative Properties
1. Boiling point
2. Freezing point
Boiling point is changed when
solute is added to the solvent.
• Boiling point elevates when a solute is
added to a solvent.
• The solution requires more energy to
reach boiling.
• Example: Salt water will not boil at
100°C. It will boil at a HIGHER temp.
Calculating the NEW Boiling Pt.
∆Tb = Kb m i
• ∆Tb = change in boiling pt.
• Kb = boiling point constant for
the solvent (will be given)
• m = molality
• i = number of ions present (USE
ONLY WITH IONIC SOLUTES!)
Example of Boiling Pt. Calculation
• What is the boiling point when 15.0g
NaCl is dissolved into 200 mL of water?
(Kb of water is 0.52 °C/m)
– ∆Tb = Kb m i
• You are solving for ∆Tb , and you have the Kb to
use.
• m =You’ll need to calculate the molality from
the info in the question. (change 15.0 g of
NaCl to moles and 200 mL to kg and plug in)
• i = Finally, is the solute ionic? YES…NaCl is
ionic and will give TWO ions when dissolved.
• What is the boiling point when 15.0g
NaCl is dissolved into 200 mL of water?
• You’ll need to calculate molality first.
– 15.0g NaCl x 1 mol NaCl = 0.256 moles NaCl
58.5 g NaCl
– 200 mL = 200 g = 0.2 kg of water
– Molality = 0.256 moles NaCl/ 0.2 kg water = 1.28m
• ∆Tb = Kb m i
• ∆Tb = (0.52 C/m)(1.28m)(2)
• ∆Tb = 1.33 C (note: This is the CHANGE in boiling
point. It does NOT answer our question.)
• New Boiling Point: 100C + 1.33C = 101.33
• We took the regular boiling point of water and added the
change since boiling point ELEVATES with a solute added.
Ten Minute Partners
Boiling Point Calculations
Kb of water = 0.52°C/m
1. What is the boiling point when 30.0g CaCl2
is dissolved into 200 mL of water?
2. The molality of the above example is similar
to the problem that we worked together a
moment ago. Explain the difference in
boiling temperature.
3. CHALLENGING: How many grams of NaCl
would need to be added to the water to
change the boiling temperature of 200 mL
to 110°C?
Answers to
Boiling Point Calculations
1.
∆Tb= (1.36m)(0.52C/m)(3) = 2.12C
100C + 2.12C = 102.12C
2. The difference is due to the difference in the
number of solute particles (ions).
3. 10C = m(0.52)(2)
m = 9.62
9.62m = mole/0.2kg
mole = 1.92 mole NaCl
112.2 grams NaCl
Freezing point is changed when
solute is added to the solvent.
• Freezing point depresses when a solute
is added to a solvent.
• The solution requires a lower temp to
reach freezing.
• Example: Salt water will not freeze at
0°C. It will freeze at a LOWER temp.
Calculating a NEW Freezing Pt.
∆Tf = Kf m i
• ∆Tf = change in freezing pt.
• Kf = freezing point constant for
the solvent (will be given)
• m = molality
• i = number of ions present (USE
ONLY WITH IONIC SOLUTES)
Example of Freezing Pt.
Calculation
• What is the freezing point when 15.0g
NaCl is dissolved into 200 mL of water?
• (Kf of water is 1.86 °C/m)
– ∆Tf = Kf m i
• You are solving for ∆Tf , and you have the Kf to
use.
• m =You’ll need to calculate the molality from
the info in the question. (change 15.0 g of
NaCl to moles and 200 mL to kg and plug in)
• i = Finally, is the solute ionic? YES…NaCl is
ionic and will give TWO ions when dissolved.
• What is the freezing point when 15.0g
NaCl is dissolved into 200 mL of water?
• You’ll need to calculate molality first.
– 15.0g NaCl x 1 mol NaCl = 0.256 moles NaCl
58.5 g NaCl
– 200 mL = 200 g = 0.2 kg of water
– Molality = 0.256 moles NaCl/ 0.2 kg water = 1.28m
• ∆Tf = Kf m i
• ∆Tf = (1.86 C/m)(1.28m)(2)
• ∆Tf = 4.76 C (note: This is the CHANGE in freezing
point. It does NOT answer our question.)
• New Freezing Point: 0C – 4.76C = -4.76C
• We took the regular freezing point of water and subtracted
the change since freezing point DEPRESSES with a solute
added.
Ten Minute Partners
Freezing Point Calculations
Kf of water = 1.86°C/m
1. What is the freezing point when 30.0g CaCl2
is dissolved into 200 mL of water?
2. The molality of the above example is similar
to the problem that we solved together.
Explain the difference in freezing
temperature.
3. CHALLENGING: How many grams of NaCl
would need to be added to the water to
change the freezing temperature of 200 mL
to -8°C?
Answers to Freezing Point
Calculations
1. ∆Tf= (0.271m)(1.86C/m)(3) =1.51C
0C - 1.51C = -1.51C
2. The difference is due to the difference in the
number of solute particles (ions).
3. 8C = m(1.86C/m)(2)
molality = 2.15
2.15 = mole/0.2kg
mole = 0.43 mole NaCl
25.14 grams NaCl
Practice Practice Practice!
• The 15-4 Practice Problems in your
practice packet are boiling
point/freezing point calculations.
• The key is posted at the front of the
classroom.
• PLEASE SOLVE MANY OF THESE
PROBLEMS! You’ll see several on
Wednesday’s test.