COLLIGATIVE PROPERTIES OF SOLUTIONS

There are some important physical properties of solution which are more directly
dependent on the concentration of solute particles. In this lesson, we will examine the
properties that are called colligative (Latin, coligare – which means “tied together”)
properties which mean, they depend on the collective effect of the concentration of solute
particles present in the solution. These properties include: (1) vapor pressure lowering, (2)
boiling point elevation, (3) freezing point depression, and (4) osmotic pressure.
Because of their direct relationship to the number of solute particles, the colligative
properties are very useful for characterizing the nature of a solute after it is dissolved in a
solvent and for determining the molar masses of substances.

Effect of solute concentration on the colligative properties of solutions

The concentration or amount of nonvolatile solute (i.e., a solute that does not have a
vapor pressure of its own) in the solution influences the colligative properties of solutions.
The result will depend on the ratio between the number of solute and solvent particles in the
solution and not on the solute's identity. However, it is necessary to consider whether the
solute is an electrolyte or a nonelectrolyte.

Effect of solute concentration on the colligative properties of solutions

1. Vapor Pressure Lowering
A direct measure of the escaping capacity of molecules is vapor pressure. A pure
liquid (solvent) can achieve equilibrium with its vapor in a closed container. And the pressure
exerted by the vapor is called vapor pressure until the equilibrium is attained. A substance
that does not have a noticeable vapor pressure is nonvolatile, whereas one that has a vapor
pressure is volatile.
If a liquid evaporates readily, a significant amount of the molecules would be present
in the gas phase, resulting in a high vapor pressure. A surface completely filled by liquid
molecules is seen on the left, some of which have evaporated and formed a vapor. On the
right, a nonvolatile solute like salt or sugar has been dissolved into the solvent, having the
effect of diluting the water. The addition of a non-volatile solute resulted in lowering of the
solvent's vapor pressure. The lowering in vapor pressure depends on the amount of
dissolved solute particles. The molecular essence of the solute is not taken into
consideration since the vapor pressure is simply a solvent's physical property and does not
undergo a chemical reaction with the solvent and does not escape into the gas phase by
itself.

Figure 1: Volatile Solvent VS Non-volatile Solute

 It should be remembered that the decrease in the vapor pressure of the solution in
this case is directly proportional to the fraction of the volatile molecules in the oil, which is
the solvent’s mole fraction. It is possible to assess this decreased vapor pressure using
Raoult’s Law (1886).

Figure 2: Relationship Between Vapor Pressure and Mole Fraction of Water

Where:

Recall from the definition of mole fraction that in a two-component solution (a solvent and a
single solute), Xsolvent = 1 – Xsolute.

Example #1: If 0.340 mol of a nonvolatile nonelectrolyte are dissolved in 3.00 mol of water,
what is the vapor pressure of the resulting solution? (The vapor pressure of pure water is
23.8 torr at 25.0 °C.)
Example #2: 210.0 g of the nonvolatile solute sucrose (C 12H22O11) is added to 485.0 g of
water at 25.0 °C. What will be the pressure of the water vapor over this solution? (The vapor
pressure of pure water is 23.8 torr at 25.0 °C.)

Example #3: Calculate the vapor pressure of water above a solution at 35.0 °C that is 1.600
m fructose, C6H12O6. (The vapor pressure of pure water at 35.0 °C is 42.2 mmHg.)
 2. Boiling Point Elevation
The addition of a non-volatile solute decreases the vapor pressure of the solution, so that the vapor
pressure of the solution is returned to a value conforming to the pure solvent, the temperature must be
increased. In fact, the temperature at which the vapor pressure is 1 atm is greater than the normal boiling
point by an amount known as the boiling point elevation.
Figure 3 below shows the phase diagram of a solution and the effect that the lowered vapor
pressure has on the boiling point of the solution compared to the solvent. In this case the sucrose solution
has a higher boiling point than the pure solvent. Since the vapor of the solution is lower, more heat must be
supplied to the solution to bring its vapor pressure up to the pressure of the external atmosphere. The

boiling point elevation is the difference in temperature between the boiling point of the pure solvent and that
of the solution.

Figure 3: The Result of Lowering the Vapor Pressure in a Solution to the Boiling Point

Figure 4: Normal Boiling Point for Water (solvent) as a Function of Molality in Several Solution Containing
Sucrose (a non-volatile solute)

For dilute solution, the elevation of the boiling point is directly proportional to the molal concentration
of the solute:
 The molal boiling point elevation constant Kb, has a specific value depending on the identity of the
solvent.

3. Freezing Point Depression

The freezing point of a substance is the temperature at which the solid and liquid forms can coexist
indefinitely, at equilibrium. Under these conditions molecules pass between the 2 phases at equal rates
because their escaping tendencies from the two phases are identical.
Figure 5 below shows the phase diagram for a pure solvent and how it changes when a solute is
added to it. The solute lowers the vapor pressure of the solvent resulting in a lower freezing point for the
solution compared to the pure solvent. The freezing point depression is the difference in temperature
between the freezing point of a pure solvent and that of a solution. On the graph, T f represents the freezing
point depression.

Figure 5: The Result of Lowering the Vapor Pressure in a Solution to the Freezing Point

If a substance is applied to a solvent such as water at a given temperature, the solute-solvent

interactions prohibit the solvent from entering the solid phase, causing the temperature to drop further until
the solution solidifies. As a result, more energy must be removed from the solution to freeze it and the
freezing point of the solution is lower than that of the pure solvent.
The degree of the freezing point depression is directly proportional to the solution's molality. Thus:

△ T f =K f m

Where:
K f – is the molal freezing point depression constant, a constant that is equal to the change
in the freezing point for a 1 molal solution of a nonvolatile molecular solute
△ T f – freezing point depression

m – molality of solute

The study of colligative properties of electrolytes requires a slightly different approach than the one
used for the colligative properties of nonelectrolytes. The explanation is because electrolytes dissociate in
solution into ions, and thus as they dissolve, one unit of an electrolyte compound splits into two or more
particles. Remember that it is the total number of solute particles that determines the colligative properties
of a solution.
+ -
For example, each unit of NaCl dissociates into two ions – Na and Cl . The colligative properties of
a 0.1 m NaCl solution should, thus, be twice as high as those of a nonelectrolyte-containing 0.1 m solution,
such as sucrose. Similarly, we would expect a 0.1 m CaCl 2 solution to depress the freezing point by three
times as much as a 0.1 m sucrose solution because each CaCl 2 produces three ions. We characterize a
quantity called the Van't Hoff factor to account for this effect, given by

actual number of particles∈solution after dissociation

i=
number of formulaunits initially dissolved ∈ solution
Thus, i should be 1 for all nonelectrolytes. For strong electrolytes such as NaCl and KNO 3, i should
be 2, and for strong electrolytes such as Na2SO4 and CaCl2, i should be 3. Consequently, the equations for
colligative properties must be modified as
∆ T b=i K b m

∆ T f =i K f m
In reality, the colligative properties of electrolyte solutions are usually smaller than anticipated
because at higher concentrations, electrostatic forces come into play and bring about the formation of ion
pairs. An ion pair is made up of one or more cations and one or more anions held together by electrostatic
forces. The presence of an ion pair reduces the number of particles in solution, causing a reduction in the
3−¿ ¿
2−¿ , PO ¿
3+¿ ,SO4 4
¿
colligative properties. Electrolytes containing multicharged ions such as Mg 2+¿, Al ¿have a greater

tendency to form ion pairs than electrolytes such as NaCl and KNO 3, which are made up of singly charged
ions.
Table 1. The van’t Hoff Factor of 0.0500 M Electrolyte Solution at 25℃

KEY EQUATIONS
Vapor Pressure of P= X P
O

Solution
Boiling Point Elevation moles of solute
∆ B =K B
kg of solution

Freezing Point △ T f =K f m
Depression
Van’t Hoff Factor actual number of particles∈solution after dissociation
i=
number of formulaunits initially dissolved ∈ solution
Colligative Properties of Nonelectrolyte and Electrolyte Solutions

Boiling Point Elevation of Nonelectrolyte Solution

Previously, you have learned that a liquid boils when its vapor pressure is equal to its atmospheric
pressure (1.01 x 105 Pa). The normal boiling point of a liquid is the temperature at which its vapor pressure
is 1.01 x 105 Pa. If the vapor pressure decreases in a nonelectrolyte solution, the boiling point of the
solution tends to increase. Therefore, to reach the 1.01 x 10 5 Pa vapor pressure, the solution must be
boiled at a temperature higher than the normal boiling point temperature of the pure solvent. The increase
in the boiling point temperature is called boiling point elevation (∆T b). To determine the boiling point
elevation of a solution, we will have to use equation 3.1.1,
∆Tb = Kbm equation 3.1.1

where Kb is the molal boiling point elevation constant and m is the molality of the solute in the
solution. Molality is used in this equation since it relates to mole fraction and, thus, particles of solute and
temperature do not affect it. On the other hand, the boiling point of a solution can be determined using
equation 3.1.2,

Tb(solution)= Tb(solvent)+ ∆Tb equation 3.1.2

where Tb(solution) is the boiling point of solution, Tb(solvent) is the boiling point of pure solvent and ∆Tb is
the boiling point elevation.

Let’s do an example!

Example 3.1A: Sucrose (C22O11H22, 342 g/mol), like many sugars, is highly soluble in water; almost 1500 g
will dissolve in 1000 g of water, giving rise to what amounts to pancake syrup. Using K b of 0.514 K/mol*Kg,
estimate the boiling point elevation of such a sugar solution.

We can solve this problem in a straightforward manner. Since all the data needed is already given,
we can just directly substitute the given values in equation 3.1.1. But take note, in equation 3.1.1, the
boiling point elevation can be solved if we have the values of molality and the molal boiling point elevation
constant and since molality is not given, then we have to calculate for it first.

Given: Molar MassC22O11H22 = 342 g/mol MassC22O11H22 = 1500 g

Volumewater = 1000 g = 1Kg Kb = 0.514 K/mol*Kg

Unknown: Boiling point elevation (∆Tb)

Equation: ∆Tb = Kbm
moles of solute
molality = equation 3.1.3
mass of solvent ∈Kg
Now that we already listed all the given data, the unknown and the equation that we will use, let’s
proceed to the solution part.
Solution: Since, we are given the molar mass and mass of the sucrose, let’s get the number of moles of
sucrose.
Mass
Number of moles = equation 3.1.4
Molar mass
1500 g
Number of moles of sucrose =
342 g/mol
Number of moles of sucrose = 4.39 mol

That’s it! Now that we have the number of moles of sucrose, we can then solve for the molality of the
solution using equation 3.1.3.
moles of solute
molality (m) =
mass of solvent ∈Kg
4.39 mol
molality (m) =
1 kg
molality (m) = 4.39 mol sucrose/kg water

Finally, we have the values of molality and the K b . We can now go to the last part of the calculation, which
is to calculate for the boiling point elevation.

∆Tb = Kbm
∆Tb = (0.514 K/mol*Kg)(4.39 mol sucrose/kg water)
∆Tb = 2.25 K

Therefore, the boiling point elevation of the sugar solution is 2.25 K.

Freezing Point Depression

Recall that only solvent vaporizes from solution, so solute molecules are left behind. Similarly, only
solvent freezes, again leaving solute molecules behind. The freezing point of a solution is the temperature
at which its vapor pressure equals that of the pure solvent, that is, when solid solvent and liquid solution are
in equilibrium. The freezing point depression (∆Tf) occurs because the vapor pressure of the solution is
always lower than that of the solvent, so the solution freezes at a lower temperature; that is, only at a lower
temperature will solvent particles leave and enter the solid at the same rate. The freezing-point depression
(∆Tf) is defined as the freezing point of the pure solvent (Tf°) minus the freezing point of the solution (Tf):

∆Tf =Tf° - Tf equation 3.1.5

Because Tf° >Tf, ∆Tf is a positive quantity. Again, ∆Tf is proportional to the concentration of the
solution:
∆Tf α m
∆Tf = Kfm equation 3.1.6
 where ∆Tf is the freezing point elevation, m is the concentration of the solute in molality units, and Kf
is the molal freezing-point depression constant. The Kf of water is 1.86 °C*Kg/mol.

Let’s practice solving for freezing point depression!

Example 3.1B: Calculate the mass of ethylene glycol (formula mass = 62.1 g/mol), an antifreeze
component, that must be added to 5.0 L of water to produce a solution that freezes at -18.3°C. Assume that
the density of the water is exactly 1 g/mL.

Given: Kf = 1.86 °C*Kg/mol Tf = -18.3 °C

Molar massethylene glycol = 62.1 g/mol Densitywater = 1 g/mL Volumewater = 5.0 L

Unknown: Mass of ethylene glycol

Solution: The approach in answering this kind of problem is the same with the previous problem. It just
differs slightly since this time, it’s the mass that is being asked.

∆Tf =Tf° - Tf

∆Tf = 0 °C – (-18.3 °C)

∆Tf = 18.3 °C

Now, we have the value for the boiling point elevation. However, our task is not yet done since it’s
the mass that we’re looking for and not the boiling point elevation. Previously, we learned that ∆T f can also
be solved using molality and Kf, as shown is equation 3.1.6. To solve for the molality, we can use this
equation and transpose it.

∆Tf = Kfm
∆T f
m=
Kf
18.3° C
m=
1.86° C∗Kg/mol
m = 9.84 mol/Kg

With this, we can now solve for the mass of the ethylene glycol by substitution.
1 Kg
Number of moles of ethylene glycol = molality (Volume of water)(density of water)( ¿
1000 g
g 1 Kg
Number of moles of ethylene glycol = 9.84 mol/Kg (5000 mL)(1 ¿( )
mL 1000 g
Number of moles of ethylene glycol = 49.2 mol

We get the mass of a substance by simply multiplying the number of moles of the substance and its
molar mass.

Massethylene glycol = 49.2 mol (62.1 g/mol)

Massethylene glycol = 3.05 x 103 g = 3.05 Kg

Therefore, the mass of ethylene glycol, an antifreeze component, is 3.05 Kg.

Colligative Properties of Electrolyte Solution

The study of colligative properties of electrolytes requires a slightly different approach than the one
used for the colligative properties of nonelectrolytes. The reason is that electrolytes dissociate into ions in
solution, and so one unit of an electrolyte compound separates into two or more particles when it dissolves.
For example, electrolyte solution such as sodium chloride dissociates into two ions – Na + and Cl-. A van’t
Hoff factor (i) must be put into the previous equations involving the colligative properties of nonelectrolytes
to account this effect. The van’t Hoff factor establishes the relationship between the number of moles of
solute dissolved and the moles of particles in a solution. This factor is given by,

number of moles∈a solution

i= equation 3.1.7
number of moles of solute dissolved

where the i value for an electrolyte with an AB empirical formula is 2, whereas for AB 2, i is 3. Thus, i
should be 1 for all nonelectrolytes. For strong electrolytes such as NaCl and KNO 3, i should be 2, and for
strong electrolytes such as Na2SO4 and CaCl2, i should be 3. Table 2.1 shows the experimentally measured
values of i and those calculated assuming complete dissociation. As you can see, the agreement is close
but not perfect, indicating that the extent of ion-pair formation in these solutions at that concentration is
appreciable.

Table3.1.1. The van’t Hoff Factor of 0.0500 M Electrolyte Solutions at 25 degrees Celsius

In calculating the colligative properties of electrolytes, such as boiling point elevation and freezing
point depression, i is integrated. The general equation now becomes,
∆Tb = iKbm equation 2.1.8
∆Tf = iKfm equation 2.1.9
Let’s practice solving!
Example 3.1C: Determine the freezing and boiling point of a solution prepared by dissolving 85.40 g of
calcium chloride (CaCl2) in 200 g of water (H2O).

Just follow the process we did in the previous examples. Start with listing all the known and
unknown data before going to the solution part.

Given: Masscalcium chloride = 85.40 g Masswater = 200 g

Molar MassCaCl2 =110.98 g/mol Kf = -1.86 ° C∗Kg/mol Kb = 0.512 ° C∗Kg/mol

Take note that CaCl2 dissociates into 3 ions, 2 ions for Chlorine and 1 ion for Calcium as given in
the reaction below.
CaCl2(aq)→Ca2+(aq) +2Cl−(aq)
Unknown: Tf and Tb

First thing we have to do is to calculate for the moles of CaCl 2. We can do this by dividing the mass of the
CaCl2 to its molar mass.
mass
Number of moles =
molar mass
85.40 g
Number of moles =
110.98 g /mol
Number of moles = 0.770 mol CaCl2

Now, we can solve for the molality of the solution using equation 2.1.3.
moles of solute
molality (m) =
mass of solvent ∈Kg
0.770 mol
molality (m) =
0.200 kg
molality (m) = 3.85 mol/Kg

Using equations 2.1.8 and 2.1.9, we can consecutively solve for ∆Tb and ∆Tf.
∆Tb = iKbm
∆Tb = 3 (0.512 ° C∗Kg/mol ¿(3.85 mol/Kg)

∆Tb = 5.91 ° C
∆Tf = iKfm
∆Tf = 3 (-1.86 ° C∗Kg/mol ¿(3.85 mol/Kg)

∆Tf = -21.48 ° C

Since the normal boiling point of water is 100 ° C , using equation 3.1.2 for boiling point and 3.1.5 for
freezing point, we must add the calculated ∆Tb to 100 ° C to get the new boiling point of the solution.
Therefore, the new boiling point is 105.91° C . On the other hand, since the normal freezing point of water is
0 ° C , the new freezing point is still -21.48° C .

Using Colligative Properties to Find Solute Molar Mass

The colligative properties of nonelectrolyte solutions provide a means of determining the molar
mass of a solute. We learned that each colligative property is proportional to solute concentrations. Thus,
by measuring the property we can determine the amount (mol) of solute particles and with a given mass,
the molar mass can be calculated. Let’s have an example!

Example 3.1D: The addition of 250 mg of naphthalene in benzene elevated the boiling point of 100 g of
benzene by 0.05°C at 1.01 x 105 Pa. Calculate the molar mass of naphthalene molecules. Benzene has a
boiling point of 80.1°C and the Kb for benzene is 2.53 °C*Kg/mol.

Given this problem, first thing you have to do is to breathe. Relax. We can solve this problem
together! Now, you have to list down all the data given and the unknown. You may follow the format below
to make your calculations neat and easier to track.

Given:
MassNaphthalene = 250 mg = 0.250 g MassBenzene = 100 g ∆ Tb = 0.05°C

Pressure = 1.01 x 105 Pa Tb = 80.1°C Kb = 2.53 °C*Kg/mol

Unknown: Molar Mass of naphthalene molecules

All givens in? Already know what we have to calculate? Good! Let’s proceed to the solution part.

Solution: With all the given data, we have to think of an equation where we can use those data given to us.
Do you think we can use equation 3.1.1 to solve for the molality of the naphthalene molecules? Absolutely!
But we can’t use the equation on its original form since we’re not looking for the ∆T b, we’re looking for the
molality. So, using the knowledge you acquire from the subject algebra, we have to transpose the equation
and get the value for molality.

∆Tb = Kbm
∆ T b K b∗m
=
Kb Kb
∆T b
m=
Kb
0.05° C
m=
2.53° C∗Kg/mol
m = 0.020 mol / Kg
 We already calculated the molality of the solution. But, we are not done yet since our final goal is to
determine the molar mass of the naphthalene molecules. We know that the solution was prepared using
0.100 kg of benzene. Now, we can find the number of moles of naphthalene in the solution.
number of moles
Molality (m) =
mass of solvent ( Kg)
Number of moles of naphthalene = molality (mass of solvent)
Number of moles of naphthalene = (0.020 mol/kg)(0.100 kg)
Number of moles of naphthalene = 1.98 x 10-3 mol

Previously, we learned that to get the molar mass, we can use equation 3.1.4.
mass
Molar mass =
number of moles
0.250 g
Molar mass =
0.00198 mol
Molar mass = 128 g/mol

The molecular formula of naphthalene is C10H8 and its molar mass is 128 g/mol.