Chapter 12 worksheet 4 (ws12.4)
Colligative Properties
A pure substance has well-defined physical properties such as freezing and boiling points. In contrast,
many properties of a solution depend on the concentration of solute. Properties of solutions that
depend on the concentration (but not on the identity) of solute are called colligative (collective)
properties. You will learn how the concentration of a solute affects four properties of a solution. As the
concentration of solute increases:
vapor pressure decreases
boiling point increases
freezing point decreases
osmotic pressure increases
We will treat these effects both qualitatively and quantitatively and you will learn why they happen. We
will discuss some important applications of these effects.
Volatile vs. non-volatile components of a solution
For each colligative property we will usually consider the case of a non-volatile solute dissolved in a
volatile solvent (a typical example is any ionic compound dissolved in water).
A volatile substance is one with a measurable vapor pressure at moderate temperatures. In
general, volatility increases as intermolecular forces decrease.
Vapor pressure reduction (Raoult’s law)
Vapor pressure is the pressure inside a closed container when a liquid is in equilibrium with its own gas
phase so the rate of evaporation is equal to the rate of condensation. When a non-volatile solute is
dissolved in a volatile solvent, the vapor pressure of the solvent is reduced. As the concentration of the
solute increases, the vapor pressure of the solvent decreases.
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1. Why does the addition of a solute reduce the vapor pressure of a volatile solvent? Entropy is the
key! Remember that one of the two driving forces for any process is an increase in entropy
(disorder). Consider the following two processes occurring in a sealed container at the same
temperature.
Process 1:
Evaporation of pure water:
Process 2: Evaporation of water from a solution:
H2O(l) → H2O(g)
H2O (solution) → H2O(g)
Clearly the entropy increases in both processes (a gas is more disordered than a liquid).
a. Which process results in a greater increase in entropy? Explain.
Since pure water is more ordered than the solution, evaporation of pure water (process 1) results in
a greater increase in entropy.
b. Which process will result in a higher equilibrium vapor pressure.
Process 1 is more favorable than process 2 due to the greater increase in entropy. Therefore,
process 1 will result in a higher vapor pressure. (Adding a solute to water, reduces the vapor
pressure!)
2. An alternative explanation for vapor pressure reduction is based on the fact that as the
concentration of a solute increases, the concentration of the water decreases. Why does a lower
water concentration result in a lower vapor pressure?
When a pure liquid is in equilibrium with its own gas phase, the rate of evaporation is equal to the
rate of condensation. The pressure of the gas is called the “equilibrium vapor pressure”. When a
solute is dissolved in the liquid, the concentration of the liquid decreases which causes the rate of
evaporation to decrease. Therefore, the rate of evaporation becomes less than the rate of
condensation and the vapor pressure drops. Over time, the rate of condensation decreases until,
eventually, the rate of evaporation and the rate of condensation are again equal. A new
equilibrium is established at a new lower vapor pressure.
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The French chemist Francois-Marie Raoult found that for sufficiently dilute solutions or for solutions
with miscible components, the vapor pressure of a solvent over a solution is directly proportional to the
mole fraction of the solvent in the solution.
Raoult’s law: P1 = X1P1o
P1o is the vapor pressure of pure solvent, X1 is the mole fraction of the solvent.
3. As the mole fraction of the solute increases, the mole fraction of the solvent ____decreases_____
and the vapor pressure of the solvent ____decreases_____. (as expected).
4. Rewrite Raoult’s law in terms of X2, the mole fraction of the solute.
P1 = (1-X2)P1o
5. According to the equation you just wrote, as the mole fraction of the solute increases, the vapor
pressure of the solvent ____decreases______. (as expected).
6. The figure below is a graph of the vapor pressure of the solvent vs. the mole fraction of the
solvent for 3 different aqueous solutions. As you move along the x-axis from left to right, is the
concentration of the solute increasing or decreasing?
Decreasing
7.
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7. A solution that obeys Raoult’s law is called an ideal solution (straight line in the graph). Ideal
solutions form when the solute and solvent are miscible. (Fill in the blank with greater than, less
than, or equal to.)
a.
In an ideal solution, the strength of the solute-solvent interactions is ___equal to_____ the
strengths of the solute-solute interactions and the solvent-solvent interactions.
b. The top curve in the graph represents a solution whose vapor pressure is higher than predicted
by Raoult’s law. In this case, the strength of the solute-solvent interactions is __less than___
the strengths of the solvent-solvent and solute-solute interactions.
c.
The bottom curve represents a solution whose vapor pressure is lower than predicted by
Raoult’s law. In this case, the strength of the solute-solvent interactions is _greater than___
the strengths of the solvent-solvent and solute-solute interactions.
8. Notice that all 3 curves converge as X1 approaches 1. Why? (This is why all sufficiently dilute
solutions obey Raoult’s law! This should remind you of ideal gases!)
In very dilute solutions there are very few solvent-solute interactions. Since we will work primarily
with dilute solutions, we will assume that they are ideal. Recall that all gases behave ideally if their
pressure (concentration) is sufficiently low because there are very few interactions among the gas
particles.
9. What is the vapor pressure of water above a solution that is 5% sucrose (by moles) at 50oC?
(Vapor pressure table is attached.) Don’t forget that Raoult’s law (as written above) is in terms
of the mole fraction of water!
Xwater = 1 - Xsucrose = 1 – 0.05 = 0.95
From vapor pressure table, at 50oC, Powater = 92.5 mm Hg
Pwater = XwaterPowater = (0.95)(92.5 mm Hg) = 87.9 mm Hg
10. The vapor pressure over a solution of urea (CH4N2O) is 291.2 mm Hg. The vapor pressure of
pure water at the same temperature is 355.1 mm Hg. Calculate the mole fraction of urea in the
solution. Don’t forget that Raoult’s law (as written above) is in terms of the mole fraction of
water!
Pwater = XwaterPowater
X water 
Pwater 291.2 mm Hg

 0.8201
0
355.1 mm Hg
Pwater
Xurea = 1 – Xwater = 1 – 0.8201 = 0.1799
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11. Ethanol and water form an ideal solution (they are miscible). Both of these substances are
volatile so they both obey Raoult’s law.
a. Which is more volatile, ethanol or water? (Think about smell and ease of evaporation.)
Ethanol (has a strong odor and evaporates easily)
b. If you have a 1:1 mixture of ethanol and water, which has a higher vapor pressure at 25oC?
Ethanol (it is more volatile)
c. If you took the vapor from above the 1:1 mixture and cooled it down so that the entire vapor
phase condensed, you would no longer have a 1:1 mixture. Which would you have more of,
ethanol or water?
Ethanol (It is enriched in the vapor phase because it is more volatile)
d. This problem describes the process of fractional distillation. Imagine taking the condensate
from part c and heating it back to 25oC and repeating the process. After several repetitions,
you would have pure ethanol! Distillation is also used aboard ships to purify water from the
ocean. In this case, you don’t have to do multiple cycles of evaporation and condensation.
Why not?
The substances dissolved in the ocean are non-volatile so the vapor phase will be pure water.
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Boiling point elevation and freezing point depression
(note: boiling point = condensation point; freezing point = melting point)
When a non-volatile solute is dissolved in a volatile solvent, the boiling point increases and the freezing
point decreases. (It’s as though the solute tries to lock the solvent in the liquid phase!)
1. Recall that a liquid boils when its vapor pressure is equal to the surrounding pressure (If the liquid
is in an open container, the surrounding pressure is the atmospheric pressure). Why does the
boiling point go up when a solute is dissolved in water?
The vapor pressure decreases so you must raise the temperature in order to get the vapor pressure
to, once again, be equal to the atmospheric pressure. Water boils at a lower temperature in Denver
than in Annapolis!
2. When pure water is heated, the temperature rises until it begins to boil and the temperature
remains constant until all the water is evaporated. When a salt solution is heated, the temperature
continues to rise even after it begins to boil. Why?
As the liquid boils off, the concentration of the solute increases so the boiling temperature
continuously increases. The temperature must rise to keep the solution boiling.
3. Freezing point / melting point is the temperature at which the liquid and solid phases are in
equilibrium. We can use a phase diagram to help us see that a solution must have a lower
freezing point than the pure solvent. (Take notes on explanation.)
The vapor pressure curve (boiling point curve; liquidvapor equilibrium) for a solution lies below that for
pure water. (This results in a higher boiling point at
each value of the pressure.) The triple point occurs
where the vapor pressure curve intersects the solidvapor equilibrium curve. Thus the solid-liquid
equilibrium curve for the solution must lie to the left of
that for pure water. Thus, the melting point for the
solution is lower (at every pressure) than for pure
water.
4.
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4. We can also understand freezing point depression by thinking about the rates of freezing and
melting. Consider pure water in equilibrium with its own solid phase (ice). At this point, the rate
of freezing is equal to the rate of melting and the temperature is 0oC. Now add a solute (like
NaCl) to the liquid phase.
a. Suddenly the rate at which the water freezes decreases. Why?
In order to freeze, water molecules must hit each other. When a solute is added, the concentration
of the water decreases so the frequency of collisions decreases.
b. What happens the rate of melting?
Nothing.
c. If the temperature remains at 0oC, all of the ice melts, why? (This is why we put salt on icy
roads!)
The rate of freezing is slower than the rate of melting.
d. What can you do to get ice to form again? Explain.
Decrease the rate of melting. You can do this by lowering the temperature. At a lower
temperature, the rate of melting decreases more than the rate of freezing.
e. When a solution freezes, the solid phase is composed of pure (or nearly pure) water. (See the
figure below. This provides another nice way to purify water from the ocean!) When you
cool pure water, it begins to freeze when the temperature reaches 0oC and the temperature
remains constant until all of the water is frozen. When you cool a solution, the temperature
continues to drop even after the water begins to freeze. Why?
This is similar to why the temperature of a boiling solution continues to rise. As the water freezes,
the concentration of the solute increases so the freezing point continuously decreases.
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The increase in boiling point and the decrease in freezing point are directly proportional to the total
molality of solute particles.
Tb = Kb m i
Tf = −Kf m i
(is positive because the boiling point increases.)
(is negative because the freezing point decreases.)
(For the mathematically inclined, see my web site for the derivation of the boiling point elevation
equation. This will show you why molality, rather than some other unit, is used here!)
m is the molality of the solute (molality = moles of solute per kilogram of solvent).
Kb is the molal boiling-point-elevation constant
Kf is the molal freezing-point-depression constant
i is the van’t Hoff factor which is the number of moles of particles that appear in solution for every
mole of solute that dissolves (more on this below).
m i = total molality of solute particles (i = 1 for all solutes except ionic compounds)
Kb and Kf are different for each solvent. For water:
Kb = 0.51 oC/m
Kf = 1.86 oC/m
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5. Which of the following solutions has the lowest freezing point and the highest boiling point?
(Simply find the solution with the highest value of mi.)
A) 0.100 m sucrose B) 0.050 m CaCl2 C) 0.050 m NaCl
A) (0.100 m)(1) = 0.100 m
B) (0.050 m)(3) = 0.150 m
C) (0.050 m)(2) = 0.100 m
B has the highest concentration of particles so it has the lowest freezing point and the highest
boiling point. A and C have the same boiling and freezing points.
6. Calculate the freezing and boiling points for a 0.100 m aqueous solution of KCl.
Tf = −Kf m i = -(1.86 oC/m)(0.100 m)(2) = -0.372 oC
Tf = 0 oC – 0.372 oC = -0.372 oC

Tb = Kb m i = (0.51 oC/m)(0.100 m)(2) = 0.102 oC
Tb = 100 oC + 0.102 oC = 100.102 oC
7. Careful measurement shows that the solution in problem 6 actually freezes at -0.344oC. The
solution behaves as though the concentration of solute particles is only 0.185 m instead of
0.200 m. Why? (Take notes on explanation.)
Ionic compounds don’t really dissociate 100% when they dissolve in water. (Another little white lie
that beginning students are told!) The ions are constantly binding to each other and dissociating.
For KCl, this equilibrium is established:
KCl (aq) = K+(aq) + Cl-(aq) (“=” represents a double arrow “⇌”)
Although the equilibrium constant for this reaction is very large, it is not infinite. Note that this is
different than the equilibrium between undissolved solid and the dissociated ions that occurs in a
saturated solution:
KCl (s) = K+(aq) + Cl-(aq)
Both of these reactions are occurring simultaneously.
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8. The table below gives the actual value of i for several solutions at various concentrations at 25oC:
Solute 0.100 m 0.0100 m 0.00100 m Limiting value (infinite dilution)
Sucrose
1.00
1.00
1.00
1.00
NaCl
1.87
1.94
1.97
2.00
MgSO4
1.21
1.53
1.82
2.00
K2SO4
2.32
2.70
2.84
3.00
a. For ionic compounds, the van’t Hoff factor depends on the concentration of the solute. As the
concentration drops, i increases and eventually reaches the limiting (or ideal) value. Why?
As the total concentration of reactants and products decreases, the equilibrium:
KCl (aq) = K+(aq) + Cl-(aq) shifts to the right (toward more moles).
b. For both NaCl and MgSO4, the limiting value (ideal value) for i is 2. At a given concentration
(except in very dilute solution) the actual value for i is lower for MgSO4 than for NaCl. Why?
The ions in magnesium sulfate are 2+ and 2- so they attract each other more strongly than do the
ions in NaCl.
Note: For most problems, we will assume that we are dealing with ideal solutions. Thus, we will
assume that solutions obey Raoult’s law and that the van’t Hoff factor is the ideal value. (In an
ideal solution of an ionic compound, ion pairs do not form!)
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Osmotic Pressure
9. In an osmotic pressure demonstration, a semi-permeable membrane that contains sugar water is
connected to a glass column and submerged into a beaker of pure water. The membrane allows
water molecules to pass through but not solute molecules. Over time, the fluid rises up the
column. This process is called osmosis. Why does the liquid rise up the column? (Think
about the rate of water flow in both directions.) Eventually, the liquid stops rising. Why?
(Something is opposing the flow of water). At the start of the demonstration, we could have
prevented the liquid from rising at all? How?
The liquid rises up the column because the concentration of water is higher outside the glass
column than it is inside of the column so water flows into the column. (Water is flowing faster into
the column than out of the column.) The liquid stops rising when the weight of the fluid in the
column counterbalances the “driving force” that is pushing the water up the column. We could
have prevented the liquid from rising in the column by applying exactly the right amount of
downward pressure on the liquid in the column. If the applied pressure was greater than the
osmotic pressure, then water would flow in the opposite direction (out of the column). This is
called “reverse osmosis”.
10. The osmotic pressure of a solution is operationally defined as the pressure that must be exerted
on the surface of the solution to prevent osmosis. What would happen if even more pressure was
applied? (This is called reverse osmosis and is commonly used to purify sea water!)
See above
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11. Two solutions with the same osmotic pressure (same concentration of solutes) are said to be
isotonic. Why must intravenous fluids be isotonic with your blood? (See picture below.)
In a hypertonic solution, the concentration of solutes is higher outside of the blood cells than inside
so water flows out of the cells causing them to become dehydrated and collapse. In a hypotonic
solution, water flows into cells and causes them to swell and explode.
Osmotic pressure is a colligative property of a solution. That is, its magnitude depends on the
concentration of dissolved particles but does not depend on the nature of the dissolved particles.
Interestingly, osmotic pressure (can be calculated using an equation that is very similar to the ideal gas
equation (see your textbook for a nice explanation of why this is true):
V nRT
or
MRT
M is the total molarity of solute particles (typically, problems will involve nonelectrolytes)
R is the ideal gas constant (0.0821 L-atm/mol-K)
T is the absolute temperature
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Using colligative properties to calculate the molar mass of a nonvolatile, non-electrolyte.
One of the most important applications of colligative properties is that they can be used to determine
molar mass. This is done as follows:
1. A known mass of a substance is dissolved in a known volume of solution or mass of solvent.
(Since you know the mass of solute, you just need to figure out the number of moles to calculate a
molar mass.)
2. Either vapor pressure, boiling point elevation, freezing point depression, or (most commonly)
osmotic pressure is measured and the appropriate equation is used to calculate the concentration
(molarity from osmotic pressure, molality from boiling or freezing point, mole fraction from
vapor pressure).
3. The number of moles of the substance in the solution is determined from the calculated molarity
and the known volume of solution (osmotic pressure) or the calculated molality and the known
mass of solvent (boiling point or freezing point).
A sample of 2.05 g of the plastic polystryrene was dissolved in enough toluene to form 100 mL of
solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25oC. Calculate the molar
mass of the polystyrene.
M = RT = [(1210 Pa)(1 atm / 101,325 Pa) ] / (0.0821 L-atm/mol-K)(298 K) = 4.88 x 10-4 M
Moles of PS = (4.88 x 10-4 moles PS / L )(0.100 L) = 4.88 x 10-5moles
Molar mass = 2.05 g PS / 4.88 x 10-5 moles PS = 4.2 x 104 g/mol
Note that this is an average molar mass since polymers are usually a mixture of molecules with
different molar masses. This is because it is difficult to control the number of monomers that
polymerize.
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This table gives the vapor pressure of PURE water at various temperatures (P1o)
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