PHASE CHEMISTRY AND
COLLIGATIVE PROPERTIES
 Phase Diagrams
 Solutions
 Solution Concentrations
 Colligative Properties
Brown et al., Chapter 10, 385 – 394, Chapter 11, 423-437
CHEM120 – Lecture Series Two : 2011/01
Phase Diagrams
Solid ⇌ Liquid
Freezing/Melting
Solid ⇌ Vapour
Sublimation/Deposition
Liquid ⇌ Vapour
Boiling/Condensing
Triple point (O) - 3 phases in equilibrium
Critical Point (C) - Above TC & PC  supercritical fluid
Ref: Brown et al., chapter 10.6, pages 391-394
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PHASE DIAGRAMS
When a gas is cooled – condensed or liquid phase – temperature
called the boiling point temperature, Tb
Boiling point temperature is pressure dependent.
Tb
P
CHEM120 – Lecture Series Two : 2011/02
Common plot of pressure vs temperature – PHASE DIAGRAM
P
LIQUID
Tb = boiling point temperature
VAPOUR
T
On the line, both phases are in equilibrium.
Gases have 3 degrees of freedom of movement; rotate, vibrate
and translate.
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Molecules lose a degree of freedom (rotation) when they move
from the gaseous state to the liquid state.
When liquid solidifies, the molecules lose their translational
motion, they are in fixed positions within the solid structure.
When we cool a liquid, the molecules become part of a rigid
structure, a solid and the temperature at which this takes
place is called the freezing point temperature.
Freezing point temperature is dependent on the pressure, if
the pressure is increased, the denser phase forms readily.
As the pressure increases, the liquid freezes sooner and the
freezing point temperature increases.
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P
SOLID
P
Tf = freezing point
temperature
SOLID LIQUID
LIQUID
VAPOUR
T
T
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Transition of a solid directly into a gas is called sublimation.
Opposite is called deposition
P
SOLID
LIQUID
X
VAPOUR
T
CHEM120 – Lecture Series Two : 2011/06
The solid/vapour and liquid/vapour lines are very important
when understanding the concept of the vapour pressure of a
substance.
Let’s take an imaginary substance with the following phase
diagram:
P
SOLID
LIQUID
0.05
VAPOUR
0
150
T
CHEM120 – Lecture Series Two : 2011/07
If we place some of our substance in the liquid phase into an
evacuated container (i.e. P ~ 0) at 150°C, on the phase diagram
we are in the region where our substance wants to be a vapour.
Therefore our substance will start to vapourise and molecules
will go from the liquid phase to the vapour phase.
As this happens, the pressure inside our container will increase
as more gas molecules will collide with the sides of the
container.
This cannot go on indefinitely, in fact, both the liquid and vapour
phases will be present at equilibrium when P = 0.05 and T =
150°C
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When the liquid and the vapour are at equilibrium then the
pressure within the container is known as the VAPOUR
PRESSURE of the liquid.
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SOLUTIONS
A solution is a homogeneous mixture of a solute in a solvent.
A homogeneous mixture is one in which the composition is
the same throughout the solution. There is also only one
phase throughout the mixture
An aqueous solution is one in which water the solvent and the
dissolved substance, the solute.
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Various types of solutions
CHEM120 – Lecture Series Two : 2011/11
SOLUTION CONCENTRATIONS
Amount of solute present in a specified amount of solution or
solvent
Expressed as
• molarity (M)
• molality (m)
• mole fraction (x)
CHEM120 – Lecture Series Two : 2011/12
Solution Concentrations
Mole Fraction
Ratio of the moles of one component to the total number of
moles present
XA =
Moles of A
Total Number of Moles
Molality
Ratio of the moles of the solute per mass of SOLVENT /kg
m=
Moles of Solute (mol)
Mass of Solvent (kg)
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Molality is useful for colligative properties.
number of moles of solute (mol)
m =
mass of solvent (kg)
EXAMPLE
The acid that is used in car batteries is 4.27 mol dm-3
aqueous sulfuric acid, which has a density of 1.25 g per
millilitre. What is the molality of the acid?
CHEM120 – Lecture Series Two : 2011/13
ANSWER
The molar concentration is 4.27 mol dm-3
For 1.00 dm3 of solution, mass = ρ × V = 1.25 g mL-1 × 1000 mL
= 1250 g
We know this solution contains 4.27 mol of H2SO4
 mass of H2SO4 = 4.27 mol × 98.12 g mol-1 = 419 g
 mass of H2O = mass of solution – mass of acid
= 1250 g – 419 g
= 831 g = 0.831 kg
The molality of the solution is therefore
m = 4.27 mol/0.831 Kg = 5.14 m
CHEM120 – Lecture Series Two : 2011/14
EXERCISE FOR THE IDLE MIND
Practice Exercise p 425; Exercise 11.29, 11.30 and 11.31 p 444.
Mole fraction is the number of moles of individual component
divided by total number of moles
Mole fraction (xA) =
molesA
total moles in solution
CHEM120 – Lecture Series Two : 2011/15
EXAMPLE
An aqueous solution of hydrochloric acid contains 36% HCl
by mass. Calculate the mole fraction of HCl in the solution.
ANSWER
In 100 g of acid, we have 36 g HCl and 64 g H2O
Moles of HCl = (36 g/36.5 g mol-1) = 0.99 mol
Moles of H2O = (64 g/18 g mol-1) = 3.6 mol
xHCl = moles HCl/total moles = 0.99/(0.99 + 3.6) = 0.22
CHEM120 – Lecture Series Two : 2011/16
EXAMPLE (Practice Exercise p 427)
A solution containing equal masses of glycerol (C3H8O3) and
water has a density of 1.10 g cm-3. Calculate (a) the molality
of glycerol, (b) the mole fraction of glycerol and (c) the
molarity of glycerol in the solution.
ANSWER
(a) For a 1000 g solution, have 500 g of glycerol and 500 g of
water.
Moles of glycerol = 500 g/92.08 g mol-1 = 5.43 mol
 Molality = 5.43 mol/0.5 Kg = 10.86 m
CHEM120 – Lecture Series Two : 2011/17
(b) Moles of water = 500 g/18.02 g mol-1 = 27.75 mol
xglycerol = 5.43 mol/(5.43 + 27.75) mol = 0.164
(c) Volume of solution = 1000 g/1.10 g cm-3 = 909.1 cm3
Molarity = moles/volume of solution
= 5.43 mol/0.9091 dm3
= 5.97 M
EXERCISE FOR THE IDLE MIND
Exercise 11.32, 11.33 and 11.37 p 444.
CHEM120 – Lecture Series Two : 2011/18
COLLIGATIVE PROPERTIES
Physical properties of solutions that depend primarily on the
number of particles present and not on their nature, e.g.
 vapour pressure lowering
 boiling point elevation
 freezing point depression
 osmosis
CHEM120 – Lecture Series Two : 2011/19
Colligative Properties Summary
COLLIGATIVE PROPERTIES
Depend on number of
particles present in solution
Molality
Vapour Pressures
Raoult's Law
P = X Po
(Daltons law
& X = 1)
Phase Changes
van't Hoff
factor
Osmotic Pressure
V = nRT
Boiling Point
Elevation
Tb = Kb m
Freezing Point
Depression
Tf = Kf m
Vapour Pressure
The pressure exerted by a vapour when it is in
dynamic equilibrium with its liquid at a fixed
temperature.
Vapour vs Gas
Vapour can be liquified by increasing
the pressure at constant temperature
but a gas cannot.
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A liquid in a closed container will establish an equilibrium
with its vapour; pressure exerted by vapour in equilibrium is
vapour pressure.
Substance with a vapour pressure is volatile.
Substance with no measurable vapour pressure is nonvolatile.
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QUESTION
Consider the following data and determine which is the more
volatile; ethanol or methanol. Explain your answer.
C2H5OH
CH3OH
T (°C)
17.7
VP (Torr)
20
T (°C)
-6
VP (Torr)
20
34.9
63.5
78.4
100
400
760
34.9
49.9
64.7
200
400
760
CHEM120 – Lecture Series Two : 2011/21
Vapour Pressure
Raoult’s Law
The vapour pressure of a component in a solution is directly
proportional to the mole fraction of that component in the
solution.
PA = XA x PoA
Where
PA – vapour pressure of A above the soln
XA – the mole fraction of component A in the soln
PoA – the vapour pressure of pure component A
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Vapour pressure of a liquid is a measure of the position of
equilibrium between the rate of evaporation and the rate of
condensation.
The lowering of the vapour pressure can be quantified by
Raoult’s Law.
Raoult’s Law: The vapour pressure of the solvent in a solution
containing a non-volatile solute is directly proportional to the
mole fraction of the solvent in the solution.
Psolution = xsolvent P*solvent
P* is the vapour pressure of the pure solvent.
CHEM120 – Lecture Series Two : 2011/22
Vapour Pressure
Raoult’s Law Assumption
Raoult’s law assumes no
intermolecular forces of
either attraction or
repulsion between the
molecules present.
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EXAMPLE
Calculate the vapour pressure (atm) of an aqueous solution at
100°C which contains 10.0 g of sucrose, C12H22O11, in 1.00 ×
102 g of water.
ANSWER
P*water = 1.00 atm at 100°C
Number of moles of water = 100 g / 18.02 g mol-1 = 5.55 mol
Number of moles of sucrose = 10.0 g / 342.3 g mol-1
= 2.92 ×10-2 mol
Mole fraction of water 5.55 / (5.55 + 2.92 × 10-2) = 0.995
Psolution = xsolvent P*solvent = 0.995 × 1.00 atm = 0.995 atm
CHEM120 – Lecture Series Two : 2011/23
Vapour Pressure
Raoult’s Law - 2 Volatile components(A & B)
The total pressure above the solution is the sum of the
individual partial pressures as calculated using Raoult’s law.
Dalton’s law of Partial pressures
Ptotal = PA + PB
Raoult’s Law
PA = XA x PoA
&
PB = XB x PoB
where
XB = 1 - XA
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EXAMPLE
Calculate by how much the vapour pressure of the solvent
changes when 40.3 g of naphthalene, C10H8, is added to 135 g
of benzene, C6H6, at 20°C. The vapour pressure of benzene at
20°C is 74.6 Torr. Consider naphthalene to be non-volatile for
this problem.
ANSWER
Psolution =
xsolvent P*solvent
(P*solvent = 74.6 Torr)
Number of moles of C6H6 = 135 g/78 g mol-1 = 1.731 mol
Number of moles of C10H8 = 40.3 g/128 g mol-1 = 0.315 mol
xsolvent = 1.731 / (1.731 + 0.315) = 1.731/2.046 = 0.846
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Psolution = (0.846)(74.6 Torr) = 63.1 Torr
Therefore, V.P. of C6H6 changes by 11.5 Torr
VP
P
solvent
1 atm
Tb solution
solution
T
Tb solvent
T
Therefore the boiling point temperature of a solvent is
elevated upon addition of a non-volatile solute
CHEM120 – Lecture Series Two : 2011/25
Boiling Point Elevation
Quantification
ΔTb = Kb x m
where
ΔTb
m
Kb
= Tb(solution) – Tb(solvent)
= molality (moles solute per mass solvent)
= molal boiling point elevation constant or
ebullioscopic constant
Note:
Kb is a constant for the solvent involved
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The degree of change in the boiling point temperature can
also be quantified by the following equation
ΔTb = Kb × m
where
m = molality
Kb = molal boiling point elevation constant
CHEM120 – Lecture Series Two : 2011/26
EXAMPLE
What is the normal boiling point temperature of a 1.45 mol
dm-3 aqueous solution of sucrose?
ANSWER
ΔTb = Kb × m
ΔTb = 0.512°C kg mol-1 × 1.45 mol Kg-1
= 0.742°C
The boiling point temperature of this solution is:
100°C + 0.742°C = 100.742°C
CHEM120 – Lecture Series Two : 2011/27
EXAMPLE
What mass of naphthalene, C10H8, must be dissolved in 422 g
of nitrobenzene to produce a solution which boils at
213.76°C at 1.00 atm?
ANSWER
The normal boiling point of nitrobenzene is 210.88°C and the
molal boiling point elevation constant is 5.24°C Kg mol-1.
From the equation:
ΔTb = Kb × m
We can calculate ΔTb:
ΔTb = (213.76 – 210.88)°C = 2.88°C
m = 2.88°C / (5.24°C Kg mol-1)(1.00) = 0.5496 mol Kg-1
CHEM120 – Lecture Series Two : 2011/28
Therefore for every Kg of nitrobenzene we add 0.5496 mol of
naphthalene
For 422 g of nitrobenzene must add:
(0.5496 mol)(0.422 Kg)/1 Kg = 0.232 mol of naphthalene
mass of naphthalene = 0.232 mol × 128.16 g mol-1
= 29.7 g
CHEM120 – Lecture Series Two : 2011/29
Freezing Point Depression
Solution
Quantification
ΔTf = Kf x m
where
ΔTf
m
Kf
P
Pure solvent
T
= Tf(solution) – Tf(solvent)
= molality (moles solute per mass solvent)
= molal freezing point depression constant
or cryoscopic constant
Note:
Kf is a constant for the solvent involved
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The addition of a non-volatile solute will decrease or
depress the freezing point temperature of a solvent.
The effect is quantified by the equation:
Tf = Kf  m
Note:
Tf is always positive
Tf = Tf (solvent) – Tf (solution)
CHEM120 – Lecture Series Two : 2011/30
EXAMPLE
Calculate the normal freezing point temperature of a 1.74 m
aqueous solution of sucrose.
ANSWER
Tf = 1.86°C Kg mol-1  1.74 mol Kg-1 = 3.24°C.
 the freezing point of the solution is (0 – 3.24°C)
= -3.24°C
CHEM120 – Lecture Series Two : 2011/31
EXAMPLE
1.20 g of an unknown organic compound was dissolved in
50.0 g of benzene. The solution had a Tf of 4.92°C. Calculate
the molar mass of the organic solute.
ANSWER
Tf = Kf (benzene) × m, Kf = 5.12°C kg mol-1
Tf of pure benzene = 5.48°C
Thus, m = (5.48 – 4.92)°C/5.12 Kg mol-1 = 0.109 m
 number of moles of solute = (0.109 mol)(0.050 Kg)/1 Kg
= 5.469  10-3 mol
Molar mass = 1.20 g/5.469 × 10-3 mol = 219 g mol-1
CHEM120 – Lecture Series Two : 2011/32
Freezing Point Depression &
Boiling Point Elevation
43
Osmotic Pressure
Osmotic Pressure
The pressure that would be applied to a solution to stop
the passage through a semipermeable membrane of
solvent molecules from the pure solvent .
ΠV = nRT OR
n
Π=
x RT = M x RT
V
where
Π = osmotic pressure
V = volume
n = moles
R = Universal Gas Constant
(0.08206 L atm K-1 mol-1
or 8.314 J K-1 mol-1)
T = temperature (K)
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Osmosis - tendency of solvent molecules to pass through a
semipermeable membrane from a more dilute to a more
concentrated solution
Osmotic pressure - the excess hydrostatic pressure on the
solution compared to pure solvent.
Reverse Osmosis - if a pressure greater than the osmotic
pressure is exerted on the solution, the solvent passes back
through the membrane to the dilute side.
When a solution containing n moles of solute in a volume V m3
is in contact with the pure solvent at a temperature T K,
ΠV = nRT
where Π is the osmotic pressure in Pa and R is the gas constant
(8.3143 J K-1 mol-1).
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The colligative properties of solutions provide a useful means
of experimentally determining molar mass. Any of the four
properties can be used.
Refer to Sample Exercises 11.12 and 11.13 p436
EXERCISE FOR THE IDLE MIND
Practice Exercise p429, 432 (2 questions), 435 and 437.
Exercise 11.44, 11.45, 11.51 and 11.54 p 445.
CHEM120 – Lecture Series Two : 2011/34