Catalyst
End
Colligative Properties Lab
Lecture 4.7 – Colligative Properties
Today’s Learning Targets
• LT. 4.11 – I can express the concentration of a
solution in terms of mass percentage, mole
fraction, molarity, and molality.
• LT 4.13 – I can explain the concept of colligative
properties and calculate the freezing point
depression and/or the boiling point elevation of a
given solution.
Solution Chemistry
• Recall,
▫ Solute - The substance in the smaller amount.
Substance being dissolved
▫ Solvent – The substance in the larger amount.
Substance that does the dissolving.
Solute + Solvent  Solution

Mass Percent
• We can determine the percent of a solute in a
solution by using the equation:
Mass of Component in Solution
Mass % =
Mass of Solution
 100
Parts per Million (ppm)
• Oftentimes, the concentration of a solute is very
low when compared to the solvent.
• We use parts per million in this case:
Mass of Component in Solution
ppm =
 10 6
Mass of Solution

Class Example
• A solution is made by dissolving 13.5 g of glucose
(C6H12O6) in 0.100 kg of water. What is the mass
percentage of solute in this solution?
Table Talk
• A 2.5 g sample of groundwater was found to
contain 5.4 μg of Zn2+. What is the concentration
of Zn2+ in parts per million?
Molarity
• Recall, the molarity of the solution can be
calculated by:
Moles of Solute
Molarity =
L of Solution

Molality
• We can also calculate the molality of a solution
by:
Moles of Solute
Molality =
kg of solvent

Colligative Properties
• Colligative Properties
are properties that depend
only on the quantity of a
substance, not the identity
of the substance.
• The following are all
colligative properties:
▫
▫
▫
▫
Vapor Pressure Lowering
Boiling – Point Elevation
Freezing Point Depression
Osmotic Pressure
Boiling Point Elevation
• When you dissolve a solute in a solution, you create
more substances for the solvent to interact with.
• This increased interaction makes it harder for the
solvent to escape, so the boiling point is increased.
• We can calculate this elevation using the equation:
Tb  iK b m
• Tb = Increase in boiling point
• i = Vant hoff factor (how many ions formed when
dissolved)
• Kb = Molal boiling point depression constant
• m = Molality of solution

Freezing Point Depression
• When you dissolve a solute in a solution, you create
more substances for the solvent to interact with.
• This increased interaction makes it harder for the
solvent to form the ordered solid arrangement.
• We can calculate this elevation using the equation:
Tf  iK f m
• Tf = Change in freezing point
• i = Vant hoff factor (how many ions formed when
dissolved)
• Kf = Molal freezing point depression constant
• m = Molality of solution

Class Example
• Automotive antifreeze consists of ethylene
glycol, CH2(OH)CH2(OH), a nonvolatile
nonelectrolyte. Calculate the boiling point and
freezing point of a 25.0 % solution of ethylene
glycol in water
Table Talk
• Calculate the freezing point of a solution
containing 0.600 kg of CHCl3 and 42.0 g of
eucalyptol (C10H18O), a fragrant substance found
in the leaves of eucalyptus trees. The typical
freezing point is CHCl3 is -63.5 oC.
Molar Mass from Colligative Properties
• From colligative properties, the molar mass of an
unknown substance can be determined if mass of
dissolved solute, mass of solvent, change in
freezing/boiling point, and K are known.
T
m
K
moles of solute
m
kg of solvent

Class Example
• A solution of an unknown nonvolatile nonelectrolyte
was prepared by dissolving 0.250 g of the substance
in 40.0 g of CCl4. The boiling point of the resultant
solution was 0.357 oC higher than that of the pure
solvent. Calculate the molar mass of the solute.
Table Talk
• Camphor (C10H16O) melts at 179.8 oC, and it has a
particularly large freezing point depression
constant, Kf = 40.0 oC/m. When 0.186 g of an
organic substance of unknown molar mass is
dissolved in 22.01 g of liquid camphor, the freezing
point of the mixture is found to be 176.7 oC. What is
the molar mass of the solute?
White Board Problems
Question 1
• Which of the following aqeuous solutions has the
highest boiling point?
a. 0.5 m NaCl
b. 0.5 m KBr
c. 0.5 m CaCl2
d. 0.5 m C6H12O6
e. 0.5 m NaNO3
Question 2
• You dissolve 2 moles of NaCl in 450. kg of H2O.
What is the freezing point depression if Kf for
this solution is 1.86 oC/m?
Question 3
• How many moles of Na2SO4 must be added to
500 mL of water to create a solution that has a 2
M concentration of Na+?
Question 4
• When 31.0 grams of a nonionic substance is
dissolved in 2.00 kg of water, the observed
freezing – point depression of the solution is
0.93 oC. If Kf is 1.86 oC/m, then what is the
molar mass of this substance?
Question 5
• How many grams of MgSO4 are in 100.0 mL of a
5 M solution?
Closing Time
• Read 13.4 and 13.5 (533 to 535 and 539) and
answer the essential questions
• There will be a formal lab report on this lab!
• Do book problems: