Chapter 4
Solutions
Solutions are homogeneous mixtures of two or more
substances.
Dissolving medium is called the solvent.
Dissolved species are called the solute.
There are three states of matter (solid, liquid, and gas)
which when mixed two at a time gives nine different kinds of
mixtures.
Seven of the possibilities can be homogeneous.
Two of the possibilities must be heterogeneous.
Seven Homogeneous Possibilities
Solute
Solvent
 Solid
Liquid
 Liquid
Liquid
 Gas
Liquid
beverages
 Liquid
Solid
 Solid
Solid
 Gas
Solid
 Gas
Gas
Solutions
Example
salt water
mixed drinks
carbonated
dental amalgams
alloys
metal pipes
air
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Two Heterogeneous Possibilities
 Solid
Gas
 Liquid
Gas
dust in air
clouds, fog
• Concentration and solubility
Ways of Expressing Concentration
All methods involve quantifying the amount of solute per
amount of solvent (or solution).
Concentration
may
be
expressed
qualitatively
or
quantitatively.
The terms dilute and concentrated are qualitative ways to
describe concentration.
 A dilute solution has a relatively small concentration of
solute.
 A concentrated solution has a relatively high
concentration of solute.
Quantitative expressions of concentration require specific
information regarding such quantities as masses, moles, or
liters of the solute, solvent, or solution.
The solution process: Polar materials dissolve only in
polar solvents (NaCI/H2O), and non - polar substances are
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soluble in non - polar solvents. This is the first rule of
solubility "like dissolves like" e.g: benzene in CCI4
Solution Concentration:
The amount of solute dissolved in a given amount of
solvent or dissolved in a given amount of solution is called
the concentration of the solution. Dilute solutions have
relatively low concentrations; concentrated solutions have
relatively high concentrations. A solution that contains as
much solute as can be dissolved is called a saturated
solution; solutions with lower concentration are called
unsaturated solutions.
Methods for expressing the solution concentration:a) Weight to weight expression.
b) Weight to volume expression.
a) Weight to weight expression;
1) Weight percent (wt%)
Weight percent is defined as: Number of grams of
solute which present in 100 gram of solution.
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mass% 
N.B
weight of solute
x 100
total weight of solution
weight fraction 
weight of solute
total weight of solution
e.g. 10% by weight glucose means:
10 gm glucose + 90 gm H2O = 100 gm solution.
% Solute = (10/100) x 100 = 10 %.
% Solvent = (90/100) x 100 = 90 %
2) Mole fraction (X):
The ratio of the number of moles of one component to the total
number of moles of all components in the mixture solution
Mole fraction is essentially self-defined. In equation form
the mole fraction (usually symbolized by X) is
Mole fraction of component , X=
moles of component
Total moles of all components
XA =
mol A
Total mol
or X A =
nA
ntotal
3) Molality
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Molality is similar to but not the same as molarity. Molality,
m, is defined by,
no moles of solute
kilograms of solvent
mass
No of moles n =
(solute)
M.Wt
Molality , m=
m=
mas
1000
(Solute)x
M.Wt
W(Solvent)
N.B
W solution = W solute + W solvent
W solvent = W solution - W solute
W solvent =dV solution - W solute
Where d = density of the solution
V = Volume of the solution
Example 4.1: What is the molality of 12.5 % solution of
glucose C6H12O6, in water? M.wt. of glucose is
180.0
Solution: 1) in 12.5 % solution 12.5 gm C6H12O6 is
dissolved in l00 gm solution.
W solvent = 100 - 12.5 = 87.5 g H2O
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2) no. of moles glucose = 12.5/180
3) m=
12.5
1000
x
=0.794 m
180
87.5
Example 4.2: What are the mole fractions of solute and
solvent in a 1.0m aqueous solution?
Solution: The molecular weight of H2O is 18.0 we find the
number of moles of water in 1000 gm of H2O.
no of moles of H2O 
1000
 55.6 mol H2O
18
A 1.0 m aqueous solution contains
n solute =1.0 mol

n H2O  55.6
n total  56.6
The mole fractions are
X solute 
n solute
1.0 mol

 0.018
n total 56.6 mol
X water = 
Solutions
n H2O  55.6
 0.982
n total 56.6
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b) Weight to volume expression:
1) Molarity (M):
• The no. of moles of solute dissolved in 1 liter of solution.
• M = No. of moles of solute per volume of solution in liter
• M = (W/M. wt) solute X (1000/V) solution
Molarity , M=
moles of solute
liters of solution
(Molarity is mol of solute per Liter of solution, not per liter
of solvent. Molarity is defined so that we can always know
how many mols of solute there are in any given amount of
solution.)
In dilute aqueous solution the molarity and molality are
about the same. However, in concentrated water solutions
and in solutions where the solvent is not water the molarity
and molality are very different.
Example 4.3: a) How many grams of concentrated nitric
acid solution should be used to prepare 250 ml of
2.0M HNO3? The concentrated acid is 70.0 %
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b) If the density of the concentrated nitric acid
solution is 1.42 g/ml. What volume should be
used? M.wt. (HNO3) =63
Solution: a) 70 gm HNO3  l00 gm solution
M
W
1000
x
M.Wt 250
Mass of pure HNO3 
2
W 1000
x
63 250
2 x 63 x 250
1000
mass of HNO3 solution = 
31.5 x 100
 45.0 gm
70
b) ml cone. NHO3 = (45/1.42) = 31.7 ml cone. HNO3
Example 4.4: An aqueous solution of acetic acid was
prepared by dissolving 164.2 gm of acetic acid in 800 ml
of the solution. If the density of the solution was 1.026
gm/ml. M. wt of acetic acid = 60
Calculate:
a) The molar concentration of the solution
b) The molality
c) The mole fraction of both the solute and the solvent
d) The mole %
e) The weight %.
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Solution:
a)
M
W
1000 164.2 x 1000
x

 3.4
M.Wt solute
Vml
60 x 800
b) d = 1.026g/ml
V = 800 ml
W solution = V x d = 800 x 1.026 = 820.8 gm
W slvent = 820.8 - 164.2 = 656.6 gm
m
mass
1000
164.2 x 1000
(Solute) x

 4.17 m
M.Wt
W(Solvent)
60 x 656.6
c) no. of acetic acid moles = 164.2 / 60 = 2.737 mole
no. of H2O moles = 656.6 / 18 = 36.44 mole
Mole fraction of acetic acid =
Mole fraction of H2O =
2.737
 0.0699
2.737  36.44
36.44
 0.9299
2.737  36.44
d) mole % acetic acid = 0.0699 x 100 = 6.99 %
mole % of H2O = 0.9299 x 100 = 92.99 %
e) percentage weight of acetic acid =
percentage weight of H2O=
Solutions
164.2 x 100
 20 %
820.8
656.6 x 100
 80 %
820.8
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Try ?
1. Five grams of NaCl is dissolved in 25.0 g of H2O. What is
the mole fraction of NaCl in the solution? (Answer =0.0580)
2. What is the mole percent NaCl in the previous problem 1
(Answer = 5.80 mol %)
3. Ten grams of ascorbic acid (vitamin C), C6H8O6, is
dissolved in enough water to make 125 ml of solution.
What is the molarity of the ascorbic acid? (Answer = 5.80
mol %)
4. What is the molality of NaCl in the solution in the previous
problem 1? (Answer = 3.42 m)
5. What is the mass percent of NaCl in the solution in the
previous problem 1? (Answer = 16.7 %)
• Principles of Solubility
Factors Affecting Solubility
The extent to which a solute dissolves in solvent depends
1. The nature of the solute.
2.
The nature of the solvent.
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3.
The temperature.
4.
The pressure (for gases).
In this section we will consider in turn the effect of each of
these factors upon solubility.
1. Solute-Solvent Interactions:
1. Intermolecular forces are an important factor in
determining solubility of a solute in a solvent. The
stronger the attraction between solute and solvent
molecules, the greater the solubility.
For example,
polar liquids tend to dissolve in polar solvents.
Favorable dipole-dipole interactions exist (solutesolute, solvent-solvent, and solute-solvent).
2. Pairs of liquids that mix in any proportions are said to
be miscible.
Example: Ethanol and water are
miscible liquids. In contrast, immiscible liquids do not
mix significantly. Example: Gasoline and water are
immiscible.
3. Consider the solubility of alcohols in water. Water and
ethanol are miscible because the broken hydrogen
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bonds in both pure liquids are re-established in the
mixture. However, not all alcohols are miscible with
water. Why? The number of carbon atoms in a chain
affects solubility. The greater the number of carbons in
the chain, the more the molecule behaves like a
hydrocarbon. Thus, the more C atoms in the alcohol,
the lower its solubility in water. Increasing the number
of –OH groups within a molecule increases its
solubility in water. The greater the number of –OH
groups along the chain, the more solute-water Hbonding is possible.
4. Generalization: “Like dissolves like”. Substances
with similar intermolecular attractive forces tend to be
soluble in one another. The more polar bonds in the
molecule, the better it dissolves in a polar solvent. The
less polar the molecule the less likely it is to dissolve in
a polar solvent and the more likely it is to dissolve in a
nonpolar solvent..
Example: Most nonpolar substances have very small water
solubilities. Petroleum, a mixture of hydrocarbons, spreads
out in a thin film on the surface of a body of water rather
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than dissolving. The mole fraction of pentane, C5H12 in a
saturated water solution is only 0.00003. Few organic
compounds that dissolve readily in water, most contain - OH
groups. Three familiar examples are methyl alcohol, ethyl
alcohol, and ethylene glycol, all of which are soluble in water in
all proportions.
H
H
C
OH
H
methyl alcohol
H
H
H
C
C
H
H
ethyl alcohol
OH
H
H
H
C
C
H
OH OH
ethylene glycol
2. Solubility and Temperatune
 Experience tells us that sugar dissolves better in warm
water than in cold water.
 As temperature increases, solubility of solids generally
increases. Sometimes solubility decreases as
temperature increases (e.g., Ce2(SO4)3).
 Gases are less soluble at higher temperatures. An
environmental application of this is thermal pollution.
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Since solubility is an equilibrium concentration, we can apply Le
Chateher"s principle in order to find out what happens when the
temperature of a saturated solution is changed. It is important to
know whether the dissolving process is
exothermic
Solute + solvent → solution + heat
or
endothermic
Solute + solvent + heat → solution
The heat of solution is defined as 𝜟H for the dissolving process,
and so is equal to H
solution
exothermic case, ΔH
- (H solute + H
solution
solvent).
Therefore, for the
is negative, and for the endothermic,
positive.
Consider now a saturated aqueous solution of potassium iodide
with excess KI(s) present. For KI, ΔH solution = 21 kJ mol -1, so we
know that KI dissolves in water with the absorption of heat and
can write the saturation equilibrium equation as
21 kJ + KI(s)  K+ + IIf we raise the temperature of the saturated KI solution, we
predict according to Le Chatelier's principle that the above
equilibrium will shift to the right (1) using up some of the
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added heat (and some of the excess solid KI) and (2)
increasing the concentration of K+ and I- ions in solution.
After equilibrium has been reestablished at a higher
temperature, the concentration of dissolved KI has become
higher, that is, the solubility of KI increases with increasing
temperature.
An example of an exothermic process is the dissolving of
lithium iodide (Lil) in water for which ΔH solution 71 kJ mol-1. We
can write the saturation equilibrium equation as
LiI (s)  Li+ + I- + 71 KJ
If
we
raise
temperature
of
the
a
saturated solution of
LiI,
the
equilibrium
shifts to the left (1)
using up some of the
added heat (and Iand
Li+
ions
in
solution) and , (2)
forming more solid Lil.
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(We observe the precipitation of some Lil out of solution.)
After equilibrium has been
reestablished at a higher
temperature,
the
concentration of dissolved
Lil is lower, so we can say
that the solubility of lithium
iodide decreases with an
increase
in
temperature.
Dissolving solids or liquids
in liquids can be either positive or negative. For aqueous
solutions it is more frequently positive; so the majority of
substances have solubilities which increase with temperature.
(This should not be used as a rule, however; there are too
many exceptions.)
When gases dissolve in liquids, ΔH is usually negative; that is,
heat is liberated. (The solvation energy usually exceeds the
energy necessary to separate the molecules in the liquid.)Thus
in the majority of cases the solubility of gas decreases with
temperature. This nearly always is true in water. Boiled water,
for example, tastes "flat," in part because dissolved air (and
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chlorine) is less soluble at the boiling point and is removed from
the water.
3. Effect of Pressure on Solubility
(solubility of gas in liquid)
Pressure changes have little or no effect on solubility of
liquids and solids in liquids.
 Liquids and solids are not compressible.
Pressure changes have large effects on the solubility of
gases in liquids.
 Sudden pressure change is why carbonated drinks fizz
when opened.
Solubility of a gas in a liquid is a function of the
pressure of the gas. The higher the pressure, the
greater the solubility.
Henry’s Law – The solubility of a gas
increases in direct proportion to its
partial pressure above the solution.
Cg  kPg
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Cg - solubility of gas
Page 93
Pg - the partial pressure of the gas
k - Henry’s law constant.
C1 C2
=
P1 P 2
An application of Henry's law: preparation of carbonated
soda.
Carbonated beverages are bottled under PCO2> 1
atm.
As the bottle is opened, PCO2 decreases and the
solubility of CO2 decreases. Therefore, bubbles of CO2
escape from solution.
Problem: At 740 torr and 20°C, nitrogen has solubility in
H2O of 0.018 g /I. At 620 torr and 20°C its solubility is 0.015
g/l.
Do these data show that nitrogen obey Henry's law or not?
Example 4.5: At 25°C oxygen gas collected over water at a
total pressure of 101 kPa is soluble to the
extent of 0.0393 g dm-3. What would its
solubility be if its partial pressure over water
were 107 kPa? The vapor pressure of water is
3.0 kPa at 25°C.
Solution: P total = PH2O + PO2
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PO2 = P total - PH2O = 101-3 = 98 kPa
∵
C1 C2
=
P1 P2
𝐶2 =
∴
0.0393 C2
=
98
107
107 𝑥 0.0393
= 0.043 𝑑𝑚−3
98
• (Solution of liquids in liquids)
1- Completely miscible liquids
2- Completely immiscible liquids H2O and aniline, H2O and
chlorobenzene
3- Partially immiscible liquids, H2O and phenol, H2O and
ether
Completely miscible liquids of binary solution
a) Ideal solution
b) Non - ideal solution
a) Ideal Solution:e.g: (n - heptane / n - hexane)
(chlorobenzene / brombenzene)
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1) The force of attraction between all molecules are
identical i.e. the attraction force is not affected by
addition of other components A - A = B-B = A - B.
2) No heat is evolved or absorbed during mixing i.e. ∆H
soln. = Zero
3) The volume of solution is the sum of volumes of the
two liquids.
4) The solution obeys RaouLt's law.
At
constant
temperature
partial
the
vapor
pressure of liquid
component
in
ideal solution is
proportional to the
mole fraction of
this constituent in
solution (Figure 1).
Solutions
Figure (1): Vapor pressure of ideal
solutions
Page 96
At constant T
PA
∝ XA
PB
∝ XB
Where PoA and PoB = vapor pressure of pure liquids A
and B
PA and PB = partial vapor pressure of liquids A and B in
solution.
PA = PoA . X A
(1)
PB = PoB . X B
(2)
∵Pt = PA + PB
Pt = PoA . X A + PoB . X B (3)
Where XA = mole fraction of A =
XB = mole fraction of B=
XA + X B = 1
nA
nt
nB
nt
 XA = 1 – XB
By substituting in (1)
Pt = PoA . X A + PoB . (1 − X A ) (4)
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Or
Pt = (PoA . - PoB . )X A + PoB . (5)
N.B Composition of solution % XA and % XB.
% A= XA x 100
% B = XB x 100
Example 4.6: Heptane (C7H16) and octane (C8H18) form
ideal solutions What is the vapor pressure at
40°C of a solution that contains 3.0 mol of
heptane and 5 mol of octane? At 40°C, the
vapor pressure of heptane is 0.121 atm and the
vapor pressure of octane is 0.041 atm.
Solution:
The total number of moles is 8.0. therefore
X heptane = 3.0/8.0 = 0.375
X octane = 5.0/8.0 = 0.625
Total = X heptane . Po heptane + X octane. Po octane
= 0.375 x 0.12 +0.625 x 0.04
= 0.045 atm + 0.026 atm.
= 0.071 atm.
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Example 4.7 : Assuming ideality, calculate the vapor
pressure of 1.0 m solution of a non - volatile, on
dissociating solute in water at 50°C. The vapor
pressure of water 50°C is 0.122 atm.
Solution : From example 2 the mole fraction of water in
1.0m solution is 0.982.
PH2O = XH2O PH2O = 0.982 x 0.122 = 0.120 atm.
Problem: At 140°C, the V.P of C6H5CI is 939.4 torr and
that of C6H5Br is 495.8 torr. Assuming that these
two liquids from an ideal solution. Find the
composition of a mixture of two liquids which boils
at 140°C under 1 atm pressure?
Example 4.8: A solution is prepared by mixing 5.81 g
acetone C3H6O, (M. wt = 58.1 g/mole) 11.9 g
chloroform (CHCI3 M.wt 119.4 g/mole). At 35°C
this solution has a total vapor pressure of 260 torr.
Is this an ideal solution? Comment? The vapor
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pressure of pure acetone and pure CHCI3 at 35°C
are 345 and 293 torr, respectively.
Solution:
n acetone =
n CHCl3=
W
5.81
 0.1 mole

58.1
M.Wt
11.9
 0.1 mole
119
nt = 0.1 + 0.1 = 0.2 mole
Xacetone = 
n acetone 0.1mol

 0.5
n total
0.2 mol
XCHCl3 = 0.5
Pt = PoA . X A + PoB . X B = 345 x 0.5 + 293 x 0.5 = 319 torr. .
∵ The observed value = 260 torr
• By comparing the 2 values, shows that the solution is not
ideal.
•
The observed value = 260 less than the expected value
= 319 this is a negative deviation from Roault's law.
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a) In liquid region (apply Rault's law).
b) Non- ideal solutions (Solutions deviate from ideal
behavior).
Negative deviation
1- The force of attraction
increase by mixing
A - A, B-B < A-B
2- The vapor pressure will
be lower than that given by
Roault's law
3- H solution :- Ve
(exothermic)
4Temperature
change
when solution is formed:
increase
5- Example: Acetone-water
Positive deviation
The force of
attraction decrease by
mixing A-A , B-B > A-B
The vapor pressure will
be higher than that given
by Raoult's law.
H
solution: + Ve
(endothermic)
Temperature
change
when solution is formed:
decrease.
Ethanol-hexane
Fig.2: Vapour pressure of Fig.3: Vapour pressure of
non-ideal
solution
(-ve non-ideal solution (+ve
deviation)
deviation)
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Fractional Distillation of Binary Miscible liquids
The separation of mixture of volatile liquids into their
components is called fractional distillation, where the
distillate containing the more volatile component and the
residue the less volatile one.
a) Ideal solutions
If a mixture of 2 liquids (A and B) form a completely
miscible ideal solution and PA > PB result in B.P. of A <
B.P of B thus on boiling:1) The Liquid A boils at lower B.P than that of liquid B.
2) The liquid A which is more volatile will be passed
from the fractionating column and the liquid B which
is less volatile returned again to the distallating
flask.
A solution of intermediate b.p. between 2 pure liquid -called
azeotropic solution
b) Non - ideal solutions (solutions that exhibit
deviations from Raoults law)
1) Non - ideal solutions with minimum boiling point:
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Fig.4: Boiling point-composition diagram
In fig 5 both the liquid and vapor curves meet at a certain
composition having a minimum boiling point, such solution
at this composition called azeotropic mixture.
•
If a solution having any other compositions is distilled,
the azeotropic mixture will distill first and the excess of
(A) or (B) will remains in the flask e.g 95 % ethanol and
5 % H2O.
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2) Non - ideal solutions with maximum boiling point:
In the B.P diagram of both liquid and vapor curves
meet at a certain composition (M) having a maximum
B.P., such solution at this composition called azeotropic
mixture.
• If a solution having any other composition is distilled, the
execs of acetone or CHCI3 will distill first leaving the
azeotropic mixture in the flask.
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• Colligative Properties of Solutions
• (Solution of solid in liquids)
• Colligative properties depend on number of solute
particles.
 Colligative properties do not depend on the kinds of
particles dissolved
• There are four colligative properties to consider:
• Vapor pressure lowering (Raoult's Law).
• Boiling point elevation.
• Freezing point depression.
• Osmotic pressure.
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 Vapor pressure lowering is the key to all four of the
colligative properties.
1. Lowering of Vapor Pressure:
 Addition of a nonvolatile solute to a solution lowers the
vapor pressure of the solution.
 The effect is simply due to fewer solvent
molecules at the solution’s surface.
 The solute molecules occupy some of the spaces
that would normally be occupied by solvent.
 Raoult’s Law models this effect in ideal solutions.
Ideal solution: one that obeys Raoult’s law.
 Real solutions show approximately ideal behavior when:
•
The solute concentration is low.
• The solute and solvent have similarly sized molecules.
• The solute and solvent have similar types of
intermolecular attractions.
 Raoult’s law breaks down when the solvent-solvent and
solute-solute intermolecular forces are much greater or
weaker than solute-solvent intermolecular forces.
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Raoult’s Law – The equilibrium vapor
pressure of the solvent over the solution is
directly proportional to the mole fraction of
the solvent in the solution
0
Psolvent  X solventPsolvent
where Psolvent  vapor pressure of solvent insolution
0
Psolvent
 vapor pressure of pure solvent
X solvent  mole fraction of solvent insolution
Lowering of vapor pressure, DPsolvent, is defined a
0
Psolvent  Psolvent
 Psolvent
0
0
 Psolvent
- ( X solvent )(Psolvent
)
0
 (1  X solvent )Psolvent
Remember that the sum of the mole fractions must equal 1.
Thus Xsolvent + Xsolute = 1, which we can substitute into our
expression.
X solute  1 - X solvent
0
Psolvent  X solute Psolvent
which is Raoult' s Law
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This graph shows how the solution’s vapor pressure is
changed by the mole fraction of the solute, which is
Raoult’s law
Examples 4.9
The vapor pressure of water is 17.5 torr at 20°C. Imagine
holding the temperature constant while adding glucose,
C6H12O6, to the water so that the resulting solution has XH2O
= 0.80 and XGlu = 0.20.
What is, the vapor pressure of water over the solution
PA  X APA0  0.80 X17.5torr  14torr
Try ?
1. Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a
density of 1.26 g/mL at 25°C. Calculate the vapor
pressure at 25°C of a solution made by adding 50.0
mL of glycerin to 500.0 mL of water. The vapor
pressure of pure water at 25°C is 23.8 torr
2. The vapor pressure of pure water at 110°C is 1070
torr. A solution of ethylene glycol and water has a
vapor pressure of 1.00 atm at 110°C. Assuming that
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Raoult's law is obeyed, what is the mole fraction of
ethylene glycol in the solution?
2. Boiling point elevation

Addition of a nonvolatile solute to a solution raises the
boiling point of the solution above that of the pure solvent

is lowered as described by Raoult’s law.
 The amount that the temperature is elevated is
determined by the number of moles of solute dissolved
in the solution.
Boiling point elevation relationship is
Tb  K b m
where : Tb  boiling point elevation
m  molal concentrat ion of solution
K b  molal boiling point elevation constant
for the solvent
Example 4.9
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What is the normal boiling point of a 2.50 m glucose,
C6H12O6, solution?
Tb  K b m
Tb  (0.512 0 C/m )( 2.50m )
Tb  1.280 C
Boiling Point of the solution = 100.00 C + 1.280 C = 101.280 C
The addition of a nonvolatile solute lowers the vapor
pressure of the solution. At any given temperature, the
vapor pressure of the solution is lower than that of the pure
liquid
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The increase in boiling point relative to that of the pure
solvent, ΔTb, is directly proportional to the number of
solute particles per mole of solvent molecules.
Molality expresses the number of moles of solute per
1000 g of solvent, which represents a fixed number of
moles of solvent
Example 4.10
Automotive antifreeze consists of ethylene glycol, C2H6O2,
a nonvolatile nonelectrolyte. Calculate the boiling point of a
25.0 mass percent solution of ethylene glycol in water.
Solution:
Boiling point = (normal b.p of solvent + ∆ T
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3. Freezing Point Depression
 Addition of a nonvolatile solute to a solution lowers the
freezing point of the solution relative to the pure
solvent.
 See previous table for a compilation of boiling point
and freezing point elevation constants.
 Relationship for freezing point depression is:
Tf  K f m
where : Tf  freezing point depression of solvent
m  molal concentrat ion of soltuion
K f  freezing point depression constant for solvent
Notice the similarity of the two relationships for freezing
point depression and boiling point elevation.
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Tf  K f m vs.Tb  K b m
 Fundamentally, freezing point depression and boiling
point elevation are the same phenomenon.
 The only differences are the size of the effect
which is reflected in the sizes of the constants, Kf
& K b.
 This is easily seen on a phase diagram for a solution.
Example 4.11: Calculate the freezing point of a 2.50 m
aqueous glucose solution.
Tf  K f m
Tf  (1.86 0 C/m )(2.50m )
Tf  4.65 0 C
Freezing Point of solution = 0.00 0 C - 4.65 0 C = - 4.65 0 C
 The size of the freezing point depression depends on
two things:
1. The size of the Kf for a given solvent, which are well
known.
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2. And the molal concentration of the solution which
depends on the number of moles of solute and the kg of
solvent.
 If Kf and kg of solvent are known, as is often the case
in an experiment, then we can determine # of moles of
solute and use it to determine the molecular weight.
Example 4.12: A 37.0 g sample of a new covalent
compound, a nonelectrolyte, was dissolved in 2.00 x 102 g
of water. The resulting solution froze at -5.58oC. What is
the molecular weight of the compound?
Tf  K f m thus the
Tf 5.580 C
m

 3.00m
Kf
1.86 0 C
In this problem there are
200 mL  0.200 kg of water.
? mol compound in 0.200 kg H2O = 3.00 m  0.200 kg
 0.600 mol compound
37 g
Thus the molar mass is
 61.7 g/mol
0.600 mol
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4. Osmotic Pressure
 Osmosis is the net flow of a solvent between two
solutions separated by a semipermeable membrane.
 The solvent passes from the lower concentration
solution into the higher concentration solution.
 Examples of semipermeable membranes include:
 cellophane and saran wrap
 skin
 cell membranes
 Osmosis is a rate controlled phenomenon.
 The solvent is passing from the dilute solution
into the concentrated solution at a faster rate than
in opposite direction, i.e. establishing an
equilibrium.
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 The osmotic pressure is the pressure exerted by a
column of the solvent in an osmosis experiment.
  MRT
where :  = osmotic pressure in atm
M = molar concentrat ion of solution 
n
V
L atm
mol K
T = absolute temperatur e
 For very dilute aqueous solutions, molarity and
R = 0.0821
molality are nearly equal.
 Mm
  mRT
for dilute aqueous solutions only
 Osmotic pressures can be very large.
 For example, a 1 M sugar solution has an
osmotic pressure of 22.4 atm or 330 p.s.i.
 Since this is a large effect, the osmotic pressure
measurements can be used to determine the
molar masses of very large molecules such as:
 Polymers
 Biomolecules like
 proteins
 Ribonucleotides
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Osmosis plays an important role in plant and animal
physiological processes; the passage of substances
through the semipermeable walls of a living cell, the
action of the kidneys, and the rising of sap in trees are
examples.
Application of osmosis
1) Revesse osmosis: When an external pressure is
applied over the solution, the solvent is forced in a
direction contrary to that normally observed. This
process called reverse osmosis is used to secure pure
water from salt water. This is used in desalination of
seawater to be suitable for drinking.
2) Isotonic solution: In the living cells, the osmotic
pressure of solution is equal to the osmotic pressure of
the cell.
e.g: NaCI (0.9%) has the same osmotic pressure as
blood.
3) Hypertonic solution: A solution of higher osmotic
pressure. In this solution red blood cells shrink. The
cells are called plasmolysed.
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4) Hypotonic solution: A solution of lower osmotic
pressure. In this solution red blood cells swells up and
burst. The cell is said to be haemolysed
Example 13: A 1.00 g sample of a biological material was
dissolved in enough water to give 1.00 x 102 mL of solution.
The osmotic pressure of the solution was 2.80 torr at 25oC.
Calculate the molarity and approximate molecular weight of
the material.
  MRT M 

RT
1 atm
? atm = 2.80 torr 
 0.00368 atm = 
760 torr
0.00368 atm
M=
 1.50  10 4 M
L atm
0.0821mol K 298 K 
?g
1.00 g
1L
4 g



6
.
67

10
mol
mol 0.100 L 1.50  10 4 M
typical of small proteins
Try ?
1. What would be the freezing point and boiling point of a
solution containing 6.50 g of ethylene glycol (C2H6O2) in
200.0 g of H2O? KfH2O = 1.86°C /m, kbH2O = 0.512°C /m
(Answer = 0.977 and 100.267oC)
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2. What are the boiling point and freezing point of a
solution prepared 2.40 g of biphenyl (C12H10) in 75.0 g of
benzene? b.p of benzene = 80.1°C f.p. of benzene =
5.5°C. (Answer = 80.626 and 4.4oC)
3. A solution prepared by dissolving 0.30 g of an unknown
nonvolatile solute in 30.0 g of CCI4 has a boiling point that is
0.392°C higher than that of pure CCI4. What is the molecular
weight of the solute? Kb = 5.02°C/m.
4. Find the osmotic pressure of blood at normal body
temperature (37°C) if blood behaves as if it were a
0.296 M solution of a nonionizing solute.
5. : An aqueous solution contains 30.0 g of a protein in l.0
L. The osmotic pressure of the solution is 0.0167 atm at
25°C. What is the approximate molecular weight of the
protein?
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