Chapter 13
Physical Properties of Colloids and Solutions
There are three types of mixtures including suspensions, colloids, and solutions. One
of the main features that differentiate between these mixtures is the size of particles
dispersed.
Suspensions
In this case, large particles are dispersed or suspended throughout a medium. For
instance, sand dispersed in a solvent like water, snow flakes dispersed in air, or
precipitated particles dispersed in a solvent are all examples of suspensions. In all
these cases, suspended particles are large enough to be seen by naked eye or an
ordinary microscope; and will settle with time. Suspended particles can be separated
from the solvent by simple filtration or through centrifugation. It is important to
indicate that the physical properties of suspensions, like boiling and freezing points as
well as vapor pressure, will be little affected by the suspended particles since the
number of these particles is extremely small as compared to the number of solvent
molecules.
Solutions
Molecules or ions are dissolved in a solvent where the molecular or ionic sizes of
solutes are very small and can not be seen by excellent microscopes. Solutes form
homogeneous mixtures with solvents where solute molecules are uniformly dispersed
in the solvent. It is important to indicate that the physical properties of solutions will
be affected by the presence of solutes.
Colloids
When the particle size of solutes is intermediate between suspensions and solutions
(1-1000 nm), the mixture formed from such solutes is called a colloidal suspension or
dispersion. Usually, in colloids, the terms dispersed phase and dispersing medium
replace the solute and solvent, respectively. Milk, smoke and mist are examples of
colloidal dispersions. Solutions containing proteins are also colloidal suspensions. An
important property of colloidal dispersions is that they are stable in the colloidal state
for a long time and they show Tyndall effect. Scattering of visible monochromatic
light (Tyndall effect) can be used to identify the existence of a colloidal suspension as
well as determine the shape of colloidal particles.
Stability of Colloidal Dispersions
For a colloidal solution to be stable, dispersed particles should be prevented from
sticking together when they collide. For a liquid in liquid colloidal solution
(emulsion), an emulsifying agent is used. An emulsifying agent is a molecule with
polar head and non polar tail. The polar head orients toward the polar constituent of
the solution (dispersion medium or dispersed phase) while the hydrophobic tail
orients toward the non polar constituent.
Surfactant
Reversed Micelle
Normal Micelle
An example of an emulsifying agent is the casein in milk, which prevents fine fat
droplets from coalescing.
Colloids of solids in liquids (sols) are stabilized by adsorption of ions on the surface
of the solid resulting in particles carrying similar charges which tend to repel each
other, thus stabilizing the solution. An example is a colloidal solution of Fe2O3.xH2O
where Fe3+ ions are adsorbed on the surface of solid Fe2O3.xH2O and prevent sticking
Colloidal solutions of solids in gas (aerosols) are stabilized by picking static charges
from air constituents, and are thus stabilized.
Destabilizing Colloids
Countering the factors stabilizing the colloid can destabilize colloids. For example,
addition of PO43- ions to the iron oxide sol will force the particles to coagulate as the
excess Fe3+ ions are now surrounded with phosphate ions which neutralize the charge
on the particles. Generally, two factors are effective in destabilizing colloids:
 Heating the solution can force charged particles to collide and stick together
when the kinetic energy is larger than the repulsion forces. Forming larger
particles (lower surface area) results in lower energy.
 Addition of a strong electrolyte will shield the charge on colloidal particles
and thus they can collide and coagulate, as in the phosphate-iron oxide case.
Types of Solutions
There are several types of solutions including:
 A solute dissolved in a liquid. The most common of which is dissolving a
solute in water and most of our study in the coming chapters will use this type
of solution.
 Dissolving a solid solute in a solid can form a solid solution. An example is
alloys; where two types of solid solutions can be identified:
 Substitutional solid solutions can be formed when an atom or molecule of a
solute takes the place of an atom or molecule in the crystal lattice of a solid.
 Interstitial solid solutions are formed when the small solute atoms or
molecules are placed within the voids or interstices of the solid. Usually,
interstitial solid solutions are dense and can be very strong, like tungsten
carbide.
Concentration Units
In addition to molarity (number of moles of solute dissolved in one liter of solution),
other units can be defined:
Mole Fraction
The number of moles of a particular component to the overall number of moles of
everything in solution is called the mole fraction of that component.
For example, a solution having 1.0 mole of acetone, 2.0 moles of ethanol, 3.0
moles of methanol, and 15.0 moles of water will have:
Xacetone = 1.0/(1.0 + 2.0 + 3.0 + 14.0) = 0.05
Xethanol = 2.0/(1.0 + 2.0 + 3.0 + 14.0) = 0.10
Xmethanol = 3.0/(1.0 + 2.0 + 3.0 + 14.0) = 0.15
Xwater = 14.0/(1.0 + 2.0 + 3.0 + 15.0) = 0.70
XT = 0.05 + 0.10 + 0.15 + 0.70 = 1.0

Mole Percent
The mole fraction multiplied by 100 is called the mole percent. In the above
example:
Mol % acetone = 0.05 x 100% = 5%
Mol % ethanol = 0.10 x 100% = 10%
Mol % methanol = 0.15 x 100% = 15%
Mol % water = 0.70 x 100% = 70%

Weight Fraction
The weight of a particular component to the total number of grams of solution is
called the weight fraction of that component.
Wcomponent = g component/g solution
For example, a solution containing 12.0 g methanol and 38.0 g water has:
wmethanol = 12.0/(12.0 + 38.0) = 0.24
wwatrer = 38.0/(12.0 + 38.0) = 0.76

Weight Percent
The weight fraction multiplied by 100 is called the weight percent. In the example
above:
Wt% Methanol = 0.24 x 100% = 24%
Wt% Water = 0.76 x 100% = 76%

Note that summation of mole or weight percent of all components will add up to
100%. Also, summation of mole or weight fractions of all components will add up
to 1.0.
Molality
Molality is defined as the number of moles of solute dissolved in one kilogram of
solvent. Therefore, the volume of one molal solution can be more or less than 1.0
L. Molality is not concerned with volume. You should be able to contrast molality
and molarity and be able to use molality in calculations as well as conversion to
other concentration units as mentioned earlier.
Example
A certain aqueous solution contains 7% ethanol (C2H5OH) by mass. Calculate the
mole fraction, mole percent, and the molality
Solution



7% by mass would mean that we have 7g ethanol in a 100g sample (and,
therefore, 93g H2O)
Molar mass of ethanol is (2 x 12.0) + (6 x 1.01) + (1 x 16.0) = 46.1g/mole
Molar mass of H2O is (2 x 1.01) + (1 x 16.0) = 18.0g/mole
Number of moles ethanol = 7/46.1 = 0.152 mol
Number of moles of water = 93/18.0 = 5.17 mol


Total moles in sample is (0.152 + 5.17) = 5.32
Xethanol = 0.152/5.32 = 0.0286
Xwater = 1 – 0.0286 = 0.9714
Mol% ethanol = 0.0286x100% = 2.9%
Mol% water = 0.9714x100% = 97.1%
Molality:

In the same 100g sample above, the ethanol 0.152 mol and the H2O weighs
93g (or 0.093 kg), therefore:
molality = 0.152mol/0.093 kg = 1.63 m
Example
A MgSO4 (FW= 120.4 g/mol) aqueous solution has a weight fraction of 0.2. What is
the molality of the solution?
Solution
A 0.2 weight fraction means that the solution contains 20 g MgSO4 and 80 g water
(0.080 kg).
Mol MgSO4 = 20 g/(120.4 g/mol) = 0.166 mol
Molality = 0.166 mol/0.080 kg = 2.08 m
Example
The mole fraction of benzene in a benzene/chloroform solution is 0.45. What is the
mole fraction of chloroform (FW = 119.4) and the weight percent of benzene (C6H6,
FW = 78.1)?
Solution
Since the solution contains benzene and chloroform only, the mole fraction of
chloroform can be calculated:
Xchloroform = 1 – 0.45 = 0.55
Assuming a total number of moles equal 1.0, then:
g C6H6 = 0.45 mol x (78.1 g/mol) = 35.1 g
g Chloroform = 0.55 mol x (119.4 g/mol) = 65.7 g
Wt% C6H6 = 35.1/(35.1 + 65.7) x 100% = 34.8%
Example
A CaCl2 (FW = 111.0 g/mol) solution is 4.57 m. Find the mole fraction of CaCl 2 and
water in solution.
Solution
4.57 m means 4.57 mole CaCl2 in 1000 g of water
Number of moles of water = 1000 g/(18.0 g/mol) = 55.6 mol
XCaCl2 = 4.57 mol/(4.57 + 55.6) = 0.0759
Xwater = 55.6/(4.57 + 55.6) = 0.9421
Example
Calculate the weight percent of formic acid solution (FW = 46.03 g/mol) that is 1.099
M. The density of the solution is 1.0115 g/mL.
Solution
Assume 1000 mL of solution. We have 1.099 mol formic acid:
g formic acid = 1.099 mol * (46.03 g/mol) = 50.95 g
Weight of solution = 1000 mL * 1.0115 g/mL = 1011.5 g
Wt% formic acid = {50.95/1011.5}* 100% = 5.001%
Energy and Disorder in the Formation of Solutions
Two factors are important to consider when explaining the solubility of substances in
liquids. These factors are:
 Entropy
A process tends to occur spontaneously (preferred) when the disorder or
randomness increase. This is why a gas confined in a container will rapidly diffuse
out when the container is open and a drop of ink will diffuse in a solution till the
concentration in all regions is the same. This is a natural tendency where solutes
tend to dissolve in order to increase randomness or entropy.
 Intermolecular Attractions
Benzene does not dissolve in water while methanol is freely soluble in water.
Why is this behavior? To answer this and similar questions, we should consult our
knowledge about intermolecular forces. First, let us have a close look at the
solubility process:
 Solvent molecules should be expanded so that a space for solutes is
created. In order to expand solvent molecules, we need to input energy
in order to overcome the intermolecular forces
 Solute molecules should also be expanded and we need to input energy
to overcome intermolecular forces between solute molecules
 The solubility of expanded solute molecules in solvent will release
energy. The released energy should, at least, be of the same magnitude
as the input energy for expanding both solute and solvent molecules so
that a solute may dissolve in a solvent.
Now, consider the case of benzene and why it does not dissolve in water. A high input
energy is needed to expand water molecules since there are the relatively strong
hydrogen bonds. It seems that these hydrogen bonds will be broken so that dispersion
forces are formed between water and benzene molecules. This is an unfavorable
process since we need to input much higher energy for solute expansion but will gain
very little as dispersion forces are formed. Therefore, benzene is not expected to
dissolve in water and the two solutions are said to be immiscible.
On the other hand, methanol dissolves freely in water since water and methanol have
same type of intermolecular attractions (hydrogen bonding) and when hydrogen bonds
in the solvent are broken, new hydrogen bonds with the solute are formed. This is a
feasible process and such solutions are very stable.
When the hydrocarbon chain becomes longer and longer, the alcohol becomes less
soluble (miscible) in water since the alcohol size becomes larger which requires
breaking several hydrogen bonds to accommodate one solute molecule; an
unfavorable process. Methanol, ethanol and propanol are all completely miscible with
water while octanol is not. On the other hand miscibility decreases from butanol to
heptanol.
It could be fairly concluded that polar solutes will dissolve in polar solvents and vice
versa. In another statement, one can say that the intermolecular forces in both solute
and solvent should be alike for the solute to dissolve in the solvent. A non polar solute
like I2 has only dispersion forces as the intermolecular forces between I2 molecules
and will thus dissolve in a non polar solvent (has dispersion forces) like CCl4 but will
not dissolve appreciably in water.
Ionic solids are highly polar and will only dissolve in very polar solvents like water.
Even polar solvents like ethanol do not have enough polarity to dissolve ionic solutes.
Ions in water become hydrated where water molecules form a layer of water around
each ion; thus decreasing the available charge, which adjacent ions can feel.
Heats of Solutions
When solutes dissolve in solvents, the solution process either releases or absorbs
energy. The amount of heat that is absorbed or released when a solute is dissolved in a
solvent is called the heat of solution (Hsoln).
Hsoln is negative when the solution process releases heat (exothermic, like solubility
of LiCl, LiI, AlCl3, and Al2(SO4)3 in water) and is a positive value when the solution
absorbs heat (as for the solution of KCl, KBr, NH4Cl, and NH4NO3 in water). Hsoln =
0 in the case where no heat is absorbed or released (Hsoln = Hsolutes, like the case of
solubility of benzene in carbon tetrachloride). The magnitude of Hsoln provides
information about relative intermolecular forces of solute, solvent, and solution.
When Hsoln = 0, the solution is referred to as an ideal solution. Two points could be
made:
 The solution process will be exothermic when the intermolecular forces
between solute-solvent molecules are stronger than solute-solute or solventsolvent molecules.
 An endothermic solution process is observed when the intermolecular forces
between solute-solute and solvent-solvent molecules are stronger than solutesolvent molecules.

The solubility process of liquids or nonionized solids in liquid solvents
involves the following steps:
 Expanding solute molecules requires an input of energy equals
Hsolute to overcome potential energy.
 Expanding solvent molecules requires an input of energy equals
Hsolvent to overcome potential energy.
 Combining expanded solute and solvent molecules will release
energy equals Hcombination, due to decrease in potential energy
(attraction occurs)
Hsoln = Hcombination – (Hsolvent +Hsolute)
This can be schematically represented by the following graphics:
E
Hsolute
Hsolvent
Hsolvent +Hsolute
Hsolvent
Hcombination
Hcombination
Hsolvent
Hsolvent +Hsolute
Hsolute
Hsolute
Hsoln
Ideal Solution Hsoln = 0

Hcombination
Hsolvent +Hsolute
Hsoln
Exothermic Process
Endothermic Process
Hcombination > (Hsolvent +Hsolute)
Hcombination < (Hsolvent +Hsolute)
For the solubility of solids in liquids, imagine a path where the solid solute is
first evaporated into gaseous ions, which requires an input of energy equals
H1 (which corresponds to potential energy needed to overcome lattice
energy). The second step will be the hydration of the ions by water molecules,
which releases energy Hhydration. Three situations can be encountered:
 When Hhydration = H1 , this is an ideal solution
 When Hhydration > H1, this is an exothermic process
 When Hhydration < H1, this is an endothermic process
Schematically, these processes can be presented as below:
E
KI(s)
H1
KI(s)
K+(g) + I-(g)
H1
Hhydration
Hhydration
H1
Hhydration
K+(g) + I-(g)
K+(g) + I-(g)
KI(s)
K+(aq) + I-(aq)
K+(aq) + I-(aq)
Ideal Solution
Hsoln = 0
K+(aq) + I-(aq)
Exothermic Process
Hhydration > H1
Endothermic Process
Hhydration < H1
Solubility and Temperature
For most solids and liquids dissolved in water, the solubility increases as the
temperature is increased. This means that the equilibrium shifts towards forming more
concentrated solutions at higher temperatures. At any temperature, a solute dissolves
to the extent where the equilibrium at that definite temperature is reached.
In the case of the solubility of gas solutes in liquids, usually the solubility decreases as
the temperature is increased. This is because the solubility of gases in liquids is
almost always exothermic. According to Le Chatelier principle solubility will thus be
inversely related to temperature.
Fractional Crystallization
Solubility of substances is affected differently with a rise in temperature. Some
substances will increase their solubility in very fast rates, like NH4NO3 while others
will respond very slowly to temperature changes, like NaCl. The differential effect of
temperature on solubility of different solutes is successfully used for the purification
of solutes by fractional crystallization. The technique involves dissolving the sample
in a small amount of a hot solvent in which the desired solute is less soluble. Upon
slowly cooling the solvent, solute crystals start to separate while impurities stay in
solution. The crystals of the solute are then collected by filtration.
Effect of Pressure on Solubility
Solubility of solids and liquids in liquid solvents are usually not affected by changes
in pressure. However, solubility of gases in liquid solvents is highly dependent on
pressure. A quantitative assessment of the effect of pressure on solubility of gases can
be clearly shown by Henry's law, which states that: The solubility of a gas (Cg) in a
liquid is directly proportional to the pressure of the gas above the solution.
C g = kg P g
kg is the Henry's law constant. This relation is useful in calculation of the solubility of
a gas at some pressure provided that the solubility at some other pressure is known:
Cg1 / Cg2 = Pg1 / Pg2
Example
At 25 oC, O2 gas collected over water at a total pressure of 1.0 atm is soluble to the
extent of 0.0393 g/L. What would its solubility be if its partial pressure over water
was 800 torr. Pwater = 23.8 torr at 25 oC
Solution
PT = PO2 + Pwater
760 torr = PO2 + 23.8
PO2 = 736 torr
Substitution in the equation Cg1 / Cg2 = Pg1 / Pg2 gives:
0.0393/CO2 = 736/800
CO2 = 0.0427 g/L
Colligative Properties of Solutions
A property that depends only on the relative amounts of solute and solvent is called a
colligative property. Colligative properties are physical properties where, depending
on the amount of solute relative to solvent, would reflect a change in some physical
properties. Physical properties which will be studied will include vapor pressure,
boiling point, melting point, as well as osmotic pressure.
 Vapor Pressures of Solutions
The presence of solutes in liquid solutions has great effects on the physical properties
of the solution. The boiling point, freezing point, as well as the vapor pressure of the
solution will vary depending on the amount and nature of dissolved solute. Raoult's
law discusses the effects of solutes on the vapor pressure of solutions provided that
these solutes are non dissociable (non ionics). Raoult,s law states the vapor pressure
of solution (Psoln) is directly proportional to the mole fraction of the solvent. This can
be represented by the relation:
Psoln = xsolvent * Posolvent
Where xsolvent is the mole fraction of the solvent and Posolvent is its vapor pressure.
Therefore, it is obvious that:
 Dissolving a nonvolatile solute will result in a decrease in the vapor pressure
of the solution.
 When the vapor pressure of the solute equals the vapor pressure of the solvent,
no change in the vapor pressure of the solution would be observed.
 When the vapor pressure of the solute is more than the vapor pressure of the
solvent, the vapor pressure of the solution would be increased.
Example
10.0 g of a paraffin (FW = 282 g/mol), a nonvolatile solute, was dissolved in 50.0 g of
benzene (C6H6, FW = 78.1 g/mol). At 53 oC, the vapor pressure of pure benzene is
300 torr. What is the vapor pressure of the solution at that temperature?
Solution
nsolvent = 50.0 g/(78.1 g/mol) = 0.640
nsolute = 10.0 g/(282 g/mol) = 0.035
xsolvent = 0.640/(0.640 + 0.035) = 0.948
Psoln = xsolvent * Posolvent
Psoln = 0.948 * 300 torr = 284 torr
Solutions with more than One Volatile Component
When solutions contain more than one volatile component, the vapor pressure of the
solution is the sum of the partial vapor pressures of all components. Assume a
situation where a solution is composed from three volatile components A, B, and C,
the vapor pressure of the solution is:
Psoln = xA * PoA + xB * PoB + xC * PoC
Example
A solution containing 50.0 g of CCl4 (FW = 153.8 g/mol) and 50.0 g of CHCl3 (FW =
119.4 g/mol). At 50 oC, if the vapor pressure of pure CCl4 and CHCl3 is 317 and 526
torr, respectively, find the partial pressure of each substance and the vapor pressure of
the solution.
Solution
nCCl4 = 50.0 g /(153.8 g/mol) = 0.325 mol
nCHCl3 = 50.0 g /(119.4 g/mol) = 0.419 mol
xCCl4 = 0.325/(0.325 + 0.419) = 0.437
xCHCl3 = 1 – 0.437 = 0.563
PCCl4 = xCCl4* PoCCl4
PCCl4 = 0.437 * 327 torr = 139 torr
PCHCl3 = xCHCl3* PoCHCl3
PCHCl3 = 0.563 * 526 torr = 296 torr
Psoln = PCCl4 + PCHCl3
Psoln = 139 torr + 296 torr = 437=5 torr
The Psoln can also be calculated directly form the relation below without calculation of
partial pressures in separate steps.
Psoln = xA * PoA + xB * PoB
Psoln = 0.437 * 317 torr + 0.563 * 526 torr = 435 torr
It should be emphasized here that Raoult's law is valid for dilute solutions only while
concentrated solutions usually show deviations from that law. Solutions obeing
Raoult's law are referred to as ideal solutions while non ideal solutions show either a
negative or positive deviations from raoult's law. Three situations can be stated:
 The situation where the intermolecular forces between solute-solute and
solvent solvent are of the same strength (Hsoln = 0). No deviations from
Raoult's law as this is an ideal solution (like benzene-carbon tetrachloride
solution).
 The situation where the intermolecular forces between solute- solvent
molecules are larger than solute-solute and solvent-solvent intermolecular
forces (Hsoln = -ve, exothermic process). Negative deviations from Raoult's
law is observed (like acetone-water solution).
 The situation where the intermolecular forces between solute-solvent
molecules are less than solute-solute and solvent-solvent intermolecular forces
(Hsoln = +ve, endothermic process). Positive deviations from Raoult's law is
observed (like ethanol-hexane solution).
Fractional Distillation
The separation of mixtures of volatile components is accomplished by a technique
called fractional distillation. The idea is simple where assuming a mixture of two
volatile components (A and B) in a solution. The solution starts to boil when the
atmospheric pressure is equal to the sum of the partial pressures of A and B.
If A is more volatile than B, we can draw a schematic between temperature and mole
fraction of B, for instance.
Temperature
Boiling
point of
B
Boiling
point of
A
Pure
A
Boiling
point of
mixture
xB
Pure
B
As the mole fraction of the less volatile component (B) is increased, the boiling point
of the mixture will also increase.
Example
Assume a mixture of A and B boils at some temperature. Find the mole fractions of A
and B provided that the vapor pressures of pure A and pure B are 1140 and 570,
respectively.
Solution
The mixture boils when the total vapor pressure of the solution equals 760 torr.
Applying Dalton’s relation for partial pressures we get:
Psoln = xA * PoA + xB * PoB
760 torr = xA * 1140 torr + (1 – xA) * 570 torr
Solving the equation for xA we get:
xA = 1/3
xB = 1 – xA = 1 – 1/3 = 2/3
Example
If 1.0 mole of A is mixed with 2.0 moles of B and, at some temperature, the resulting
solution boils. Find the partial pressure of A and B provided that the vapor pressures
of pure A and pure B are 1140 and 570, respectively.
Solution
XA = 1.0/(1.0 + 2.0) = 1/3
XB = 2.0/(1.0 + 2.0) = 2/3
PA = x A * PoA
PA = 1/3 * 1140 torr = 380 torr
PB = xB * PoB
PA = 2/3 * 570 torr = 380 torr
For the composition of the vapor, we may write:
PA = xA(vapor) PT(vapor)
380 torr = xA(vapor) * 760 torr
xA(vapor) = 0.500
Note that the mole fraction of A in solution is just 0.333, it was increased in the vapor.
Also, we can calculate xB(vapor) to be also 0.500
Therefore, it can be concluded that the vapor will be richer in the more volatile
component. This can be easily seen from studying the boiling point diagram below:
Vapor
composition
curve
Temperature
T1
T2
Liquid boiling point curve
Pure
A
X3
X2
xB
X1
Pure
B
The upper curve on the boiling point diagram indicates the composition of the vapor.
The horizontal line between the liquid boiling point curve and the vapor composition
curve is called a tie line. To understand the diagram, we read it as follows:
 When the composition of the mixture is x1, it boils at temperature T1, to
provide a vapor that has a composition x2. The vapor is collected.
 When the vapor with composition x2 is condensed and reheated, it will boil at
temperature T2.
 When the vapor with composition x2 boils at T2, the vapor composition will be
x3 .
 Repetition of this process will result in vapor that is continuously richer in
component A.
This procedure is a typical fractional distillation technique in which volatile
liquids are separated from each other.
Solutions that Show Large Deviations from Raoult’s Law
There are some solutions that show very large deviations from Raoult’s law.
Deviations can be positive or negative where positive deviations (higher vapor
pressures) mean the presence of a low boiling composition like the one in the
figure below. A solution with such a minimum is called a minimum-boiling
azeotrope. Fractional distillation of solution compositions below or above the
minimum is possible and will give either component plus the azeotrope. However,
separation of the mixture at the azeotrope composition is not possible with
fractional distillation. Therefore, in terms of boiling point the azeotrope behaves
as if it were a new substance.
Low Boiling Azeotrope (like propanolwater mixture)
High Boiling Azeotrope (like aquous
HCl solution)
 Freezing Point Depression
Since nonvolatile solutes lower the vapor pressure of solutions, the freezing point of
the solution will also be affected where depression of the freezing point is observed.
This is a consequence of the effect of the nonvolatile solute on the properties of
solutions. The phenomenon is clear when the phase diagram for the solution is
studied:
From the graph, as the vapor pressure decreases the phase transfer from liquid to solid
would occur at a lower temperature. The decrease in freezing point depends on the
depression in vapor, due to presence of the nonvolatile solute. The relation governing
the effect of amount of added nonvolatile solute on freezing point depression is:
Tf = kf m
Where Tf is the freezing point depression, kf is a molal freezing point depression
constant and m is the molality of the solution.
 Boiling Point Elevation
In addition to depression in freezing point, it should be obvious that the boiling point
of a solution having a nonvolatile solute should be higher than the pure solvent. This
is simply due to the lower vapor pressure of the solution where boiling point would be
reached when the vapor pressure of solution equals the atmospheric pressure. Since
the vapor pressure is lowered as a result of the nonvolatile solute, the boiling point is
increased. This is very clear from the phase diagram above. The relation governing
the effect of amount of added nonvolatile solute on boiling point elevation is:
Tb = kb m
Where Tb is the boiling point elevation, kb is a molal boiling point elevation constant
and m is the molality of the solution.
Example
What is the freezing point and boiling point of a solution containing 6.50 g of
ethylene glycol (FW = 62.1 g/mol) in 200 g of water? Kf = 1.86 oC/m and kb = 0.51
o
C/m
Solution
m = mol solute/kg solvent
mol solute = Wt/FW = 6.50g/(62.1 g/mol) = 0.105 mol
Tf = kf m
Tf = 1.86 oC/m * {0.105 mol/0.200 kg} = 0.976 oC
Tb = kb m
Tb = 0.51 oC/m * {0.105 mol/0.200 kg} = 0.27 oC
Therefore, the solution will boil at 100.27 oC and will freeze at – 0.976 oC.
An important application of the phenomena of boiling point elevation and freezing
point depression is the determination of molecular masses of nonvolatile solutes; as in
the example below:
Example
A 5.50 g of a newly synthesized compound was dissolved in 250 g of benzene (kf =
5.12 oC/m) and the freezing point depression was found to be 1.2 oC. Find the
molecular mass of the compound.
Solution
Tf = kf m
1.2 oC = 5.12 oC/m * m
Molality = 0.199 m
Molality = mol solute/kg solvent
0.199 = mol solute/0.250 kg
mol solute = 0.0498 moles
mol solute = Wt solute/FW
FW = 5.50 g/0.0498 mol = 110 g/mol
It should be realized that a solvent with high kf is an advantage for such experimental
calculations of molecular masses.
 Osmotic Pressure
Osmosis is defined as the process whereby solvent molecules pass from a dilute
solution to a more concentrated one, through a semi permeable membrane. An
equilibrium force which will ultimately bring both solutions to the same concentration
drives this process forward. An osmometer is a device used for measurement of
osmotic pressure (). The solution is placed in the osmometer, which is inserted in a
solvent. Solvent molecules pass through the osmotic membrane and the value of the
height of the solution in a capillary is measured and related to the osmotic pressure. A
relation for the osmotic pressure can be shown to resemble the ideal gas equation
where:
is proportional to the molarity of the solution as well as the absolute temperature,
where:
= MRT
Where R is the gas constant
M = n/V where M is the molarity of the solution
= (n/V)RT
V= nRT
This is called the van’t Hoff equation and is very similar to the common ideal gas law
(PV = nRT).
The magnitude of the osmotic pressure, even for dilute solutions, is usually large and
thus can be advantageously used for accurate determination of molecular masses.
Solutions that have the same osmotic pressure are called isotonic solutions.
Reverse Osmosis
If a pressure greater than the osmotic pressure is applied at the surface of the solution,
solvent will pass from the more concentrated to dilute solution. This process is called
reverse osmosis and is the basis for a wide spreading technique for water purification.
Solutions of Electrolytes
All previous discussion about effect of nonvolatile solute on colligative properties
was limited to non-electrolyte solutes. Electrolytes, on the other hand, are even more
effective in causing lowering in vapor pressure and freezing point and more elevation
in boiling point of solutions. This is simply because when electrolytes dissociate in
solution, they produce more species. For instance, 1.0 m NaCl will produce 1.0 m Na+
and 1.0 m Cl- and the overall effective molality of the solute will be 2.0 m.
Example
What is the freezing point depression of a 0.15 m aqueous solution of Al2(SO4)3? (kf =
1.86 o C/m)
Solution
Al2(SO4)3  2 Al3+ + 3 SO42-
Therefore, the overall effective molality of the solution is:
m = 0.15 * 5 = 0.75
Tf = {1.86 oC/m}*0.75 m = 1.4 oC
Example
A solution of CaCl2 (FW = 111 g/mol) was prepared by dissolving 25.0 g of CaCl2 in
exactly 500 g of H2O. What is the expected vapor pressure at 80 oC (Pwater at 80 oC =
355 torr)? What would the vapor pressure of the solution be if CaCl2 were not an
electrolyte?
Solution
Psoln = xsolvent Posolvent
Therefore, calculate the number of moles of water and CaCl2
nwater = 500g/{18.0 g/mol} = 27.8
nCaCl2 = 25.0 g/{111 g/mol} = 0.225
nspecies = 3 * 0.225 = 0.675
xwater = 27.8/{27.8 + 0.675} = 0.975
Psoln = 0.975 * 355 torr = 346 torr
If CaCl2 were not an electrolyte, CaCl2 will not dissociate and the number of moles of
all species of solute will be 0.225:
Xwater = 27.8/{27.8 + 0.225} = 0.993
Psoln = 0.993 * 355 torr = 352 torr
It is clear from this example that the pressure will decrease if the solute is an
electrolyte. In fact, electrolytic solutes will result in increased depression in vapor
pressure and freezing point and increased elevation in boiling point, mainly because
electrolytes dissociate in solutions.
Interionic Attractions and the van’t Hoff Factor
The above examples assume that positive and negative ions have no attraction forces
and thus each ion behaves as an intact species. This is, in fact, not true since there are
some attractions which result in decreasing the number of independent species. For
example, the following table provides some quantitative assessment of attractions:
Salt
0.1 m
0.01 m
0.001 m
NaCl
K2SO4
MgSO4
1.87
2.32
1.21
1.94
2.70
1.53
1.97
2.84
1.82
Theoretical
factor (i)
2
3
2
Note that as the electrolytic solution becomes more dilute, the effect of attraction
forces is diminished. Ultimately, Attraction forces will be zero at infinite dilution. The
degree to which an electrolyte behaves as if its ions were independent is quantitatively
given by van’t Hoff factor, i:
i = Tf (measured)/{Tf calculated as non electrolyte}
It should also be observed that attractive forces are even more effective for multiply
charged ions.
Example
Find the boiling point elevation of a 0.100 m MgSO4 aqueous solution (kb = 0.51
o
C/m, i = 1.21).
Solution
Tb = kb * m
m = 1.21 * 0.100 m = 0.121
Tb = 0.51 oC/m * 0.121m = 0.062 oC
If MgSO4 was not an electrolyte
Tb = 0.51 oC/m * 0.100 m = 0.051 oC.