Honors Chemistry
Name: _______________________________________ Date: _______________ Mods: __________
Ch. 13.5 – Colligative Properties

Colligative Properties:
o Properties of a solution which depend on the ________________________ of solute
particles but NOT the ________________________ of the solute are called colligative
properties.
 Vapor pressure lowering (VP ↓)
 Boiling point elevation (BP ↑)
 Freezing point depression (FP ↓)
 Osmotic pressure

Ionic vs. Molecular Solute:
o
When it comes to colligative properties, we are comparing VP, BP, FP, and osmotic
pressure of a pure solvent to a solution made by dissolving a solute in that solvent
 I WILL SAY THIS AGAIN BECAUSE IT IS SO IMPORTANT: Colligative
properties depend ENTIRELY on the # of solute particles dissolved in
solution (the greater the # of solute particles dissolved, the greater the
impact on colligative properties)
o
Soluble __________________ compounds:
 Solute dissolves in a solvent to produce multiple ions (particles)
 All strong electrolytes will do this (including strong acids)
o
Soluble _________________________________ compounds
 Solute dissolves in a solvent but remains as one intact molecule
(only produces 1 particle when dissolved)
 These are nonelectrolytes because there are no ions present once dissolved
o
Because ionic compounds dissolve and form a greater # of particles in solution, ionic
solute have a ________________ significant impact on colligative properties than
molecular solutes do.
Aqueous
Solutions
(solute dissolved
in water)
1.5 m CH3OH
1.5 m NaCl
1.5 m Na2SO4
Ionic or
Molecular?
List of particles/ions
dissolved in solution
# of
particles in
solution
Impact on
colligative
properties

Vapor Pressure Lowering
o
Vapor pressure is _______________ proportional to the
#of gas molecules above a liquid or solution.
o
When a solute is dissolved into a solvent, the vapor
pressure is affected by the # of solute particles:
 The greater the concentration of solute particles, the
_____________________ the vapor pressure above the solution
 Why? – more solute particles dissolved means _________________________
solute-solvent IMFs in the solution ... this means less gas particles are able to
escape from the solution = lower VP above the solution
o
Raoult’s Law: allows us to calculate the actual vapor pressure (Psolvent) above any
solution at a specific temperature.
 Psolvent = is the partial vapor pressure exerted by the solvent above a solution at
that temperature
 Xsolvent = the mole fraction of the solvent
 P° solvent = the normal vapor pressure of pure solvent at that temperature
NOTE: You need to make sure you have all variables in regards to the __________________
Example Problem #1:Calculate the vapor pressure of water above a solution prepared by adding 22.5
g of lactose (C12H22O11) to 200.0 g of water at 338 K. How much was the vapor pressure lowered after
the addition of the lactose solute? (Note: the vapor pressure of pure water at this temperature is 187.5
torr)

Boiling Point Elevation
o
Adding a solute to a pure solvent will ____________________ (or elevate) the
temperature at which the solution boils
 Ex) adding salt to water when boiling pasta (pure water boils at 100ᵒC, salt
water has a BP higher than 100ᵒC)
o
Why is this so?



Boiling occurs when atmospheric pressure and vapor pressure are in
equilibrium (both are at the same pressure)
Adding solute means stronger IMFs in the solution (compared to the pure
solvent); stronger IMFs means less VP at the original BP of the pure solvent,
which means a hotter temperature (higher BP) is required in order for enough
gas molecules to escape to the vapor phase and raise the VP high enough to
be equal to the atmospheric pressure (which doesn’t change)
Freezing Point Depression
o
Adding a solute to a pure solvent will _______________________(or depress) the
temperature at which the solution
freezes and melts

o
Ex) adding salt to icy
roads (pure ice melts at
0ᵒC but salty ice can melt
below 0ᵒC); addition of
ethylene glycol
(antifreeze) to car
radiators.
Why is this so?
 In order to freeze a
solution, the molecules
must crystallize and come
together in a crystal
lattice structure. An

added solute interferes with the crystallization so a lower temperature is
required to freeze a solution.
Boiling Point Elevation & Freezing Point Depression Calculations
o
The change in the boiling point of a solution is directly proportional to the
________________________ (m) of the solution
ΔTb =
 i = # of particles/ions in solution (van’t hoff factor)
 kb = “molal boiling point elevation constant” a property of the solvent
 ΔTb must be _____________________ to the normal boiling point of the pure
solvent to determine the actual boiling point of the solution
o
The change in freezing point can be found similarly:
ΔTf =
 i = # of particles/ions in solution (van’t hoff factor)
 ΔTf must be _________________________ from the normal freezing point of
the solvent to determine the actual freezing/melting point of the solution

Colligative properties and electrolytes
o
In a problem such as freezing point depression or boiling point elevation, a 1.0 m
aqueous solution of NaCl (2 ions in solution) the equation to solve for BP elevation
would be:
Tb = i  kb  m = (2)(0.51ºC/m)(1 m)
… this will make a big difference when calculating with colligative properties…
 Therefore, you must know if the solute is an electrolyte (i = # of ions) or
a nonelectrolyte (i = 1)
Example Problem #2: Rank the following in order of increasing freezing point, meaning from the
lowest to highest FP temp.) (Note: the greater the molality and the # of particles the greater ΔTf)
a) 0.050 m CaCl2
b) 0.15 m NaCl
c) 0.10 m HCl
d) 0.10 m C12H22O11
Example Problem #3: Automotive antifreeze consists of ethylene glycol (C2H6O2). Calculate the
boiling point and freezing point of 25% by mass solution of ethylene glycol in water. (Note: for water
the kb = 0.51 °C/m and the kf = 1.86 °C/m)
Example Problem #4: A solution of some unknown covalent solute was prepared by dissolving 0.250
g of the substance into 40.0 g of CCl4. The boiling point of the resultant solution was 0.357ºC higher
that the pure solvent. Calculate the molar mass of the solute.

Osmosis
o
Some substances form _____________________________ membranes

allows some smaller particles to pass through, but blocks out larger particles.
o
In biological systems (plants, animals, humans), most semipermeable membranes
allow ___________________ molecules (water) to pass through, but
___________________ molecules cannot pass.
o
In osmosis, the ______________________ moves through the semipermeable
membrane from a region of ____________ solute concentration to a region of
______________ solute concentration

This is because it is trying to achieve _______________________ by having
the same concentration (aka: isotonic) on both sides of the membrane

Sometime equal concentrations on both sides of the membrane are not
possible, however water will stop flowing when a great enough pressure is
exerted on the membrane from the side of higher concentration.
o
Osmosis in blood cells:
 If the solute concentration outside the cell is _________________ than inside
the cell, the solution is __________________________. Water will flow out of
the cell and it will shrivel.
 If the solute concentration outside the cell is ______________ than inside the
cell, the solution is __________________________. Water will floe into the cell
eventually causing it to rupture.
Hypertonic

Hypotonic
Osmotic Pressure
o
The applied pressure needed to stop osmosis from occurring is known as osmotic
pressure (π)
π=






I = # of solute particles
n = # of moles
V = volume of solution
M = molarity of solution
R = ideal gas constant
T = temperature (in K)
Example Problem #5: At 31 °C, what will be the osmotic pressure of a 0.21 M NaCl solution?
Colligative Properties: Calculations WS
1. Indicate how many particles are formed when the following solutes dissolve.
SOLUTE
2.
# OF PARTICLES
SOLUTE
sucrose (C12H22O11)
magnesium chloride (MgCl2)
sodium sulfate (Na2SO4)
methanol (CH3OH)
# OF PARTICLES
Arrange the following solutions in order of decreasing boiling point (highest BP to lowest BP):
a) 0.15m CH3COOH
b) 0.1m H2SO4
c) 0.1m Mg(NO3)2
d) 0.1 m NaBr
3. Which salt, KBr or CaCl2, has a greater effect on freezing point? Explain.
4. Raoult’s Law: Psolvent (over solution) = (Xsolvent)(P°solvent)
12 g Sucrose (C12H22O11) is dissolved in 250.0 g of water at 90°C. At this temperature, pure H2O
has a partial pressure (P°) of 525.8 mm Hg. What is the vapor pressure of water over this
solution?
5. Freezing Point Depression: ΔTf =i ● kf ● m
A solution is prepared by adding 0.50 g of caffeine (C8H10O2N4) to 100 g of benzene (C6H6).
Calculate the freezing point of this solution. (Note: the freezing point of pure benzene is 5.50°C
and the kf for benzene = 5.23 °C/m)
6. Boiling Point Elevation: ΔTb =i ● kb ● m
An aqueous solution is prepared by dissolving some potassium phosphate (K3PO4) in 500 g of
water. The normal boiling point of this potassium phosphate solution is 100.92°C. What mass of
K3PO4 was dissolved to make this solution? (Note: kb for water = 0.51°C/m … you should know
the boiling point of pure water )
7. Salt is often used to remove ice from roads and sidewalks. Explain how this process works in
terms of colligative properties.
8.
What is the freezing point of an aqueous sucrose solution that has 25.0 g C12H22O11 per 100g
H2O? (Note: kf for water = 1.86°C/m … you should know the freezing point of pure water )
9.
Determine the freezing point depression (ΔTf) of a 0.25M solution of NaCl, given the
assumption that the density of this solution is 1.1 g/ml. (Note: kf for water = 1.86°C/m)
10. What mass of lithium sulfate (Li2SO4) should be added to a 75.0 g of H2O to raise the boiling
point of the solution to 100.3°C? (Note: kb for water = 0.51°C/m … you should know the boiling
point of pure water )
11. When 5.0 g of CaCl2 dissolves in 50.0 g of water, what is the boiling point of the solution? (Note:
kb for water = 0.51°C/m … you should know the boiling point of pure water )
12. Find the boiling point of a solution containing 6.0 g benzene, C6H6, in 35 g of chloroform, CHCl3.
(Note: kb for CHCl3= 3.6°C/m and the boiling point is 61.2ᵒC
13. Mrs. J’s feet are aching at the end of a long day. At home, she dissolves 26.0 g of Epsom salt,
MgSO4, in 1.5 kg of water. What is the freezing point of this solution? (Note: kf for water =
1.86°C/m … you should know the freezing point of pure water )
14. Osmotic Pressure: π = iMRT
What is the osmotic pressure of a 0.1 M solution of Ca(NO3)2 at 25°C?