The document discusses liquid solutions and Raoult's law, which describes the vapor pressure of volatile components in a binary solution. It explains deviations from Raoult's law, leading to positive and negative deviations, and highlights the concept of azeotropes, which are mixtures that boil at a constant temperature. Additionally, it covers the behavior of solid solutions, emphasizing how the presence of non-volatile solutes affects the vapor pressure and boiling point of solvents.

1. SOLUTIONS
LIQUID IN LIQUID
RAOULT’S LAW

2. Liquid Solutions
● Liquid solutions are formed when solvent is a liquid.
● The solute can be a gas, a liquid or a solid.
● In this Section, we shall discuss the solutions of liquids and solids in a
liquid.
● Such solutions may contain one or more volatile components.
● Generally, the liquid solvent is volatile.
● The solute may or may not be volatile.

4. P1
0
= Vapour pressure of pure component
P1= Vapour pressure of component 1 after
mixing with component 2

5. Let us consider a binary solution of 2 volatile liquids 1 and 2. When taken in a
closed system, both the liquids will evaporate and soon a equilibrium will be
established between the liquid phase and vapour phase.
Under this condition,
P total = total vapour pressure
P1 = partial vapour pressure of component 1
P2 =partial vapour pressure of component 2
X1 = mole fraction of component 1
X 2 = mole fraction of component 2
p10 = partial vapour pressure of pure component 1
p 0 = partial vapour pressure of pure component 2

6. “For a solution of volatile liquids the partial vapour pressure of each component
is directly proportional to its mole fraction.”
Thus for component 1
P1 x
∝ 1 ⇒⇒⇒ p1= p1
0
x1
Similarly for component2
P2 x
∝ 2 ⇒ p2=p2
0
x2
The French scientist Raoult gave a quantitiative relation between between them
which is known as Raoult’s Law.
Where p1
0
= vapour pressure of pure component 1
And p2
0
= vapour pressure of pure component 2

7. Using this relation the the total VP can be calculated if mole fraction of
one component is known.
If a graph is plot between the mole fraction of component 1 and 2 Vs the
vapour pressure of the components we come across the following
observations.

8. MOLE FRACTION
V
A
P
O
U
R
P
R
E
S
S
U
R
E
X1= 1
X2= 0
X1= 0
X2= 1
Ptot = P1+P2
P2
=P2
0 .x2
P1 =P1
0
.x1
P1
0
P2
0
X1= 0.7
X2= 0.3

9. Assuming that component 1 is less
volatile than 2 i.e p1
0
<p2
0
➢ Total VP over the solution varies
directly with the mole fraction
component 2.
➢ The total vapour pressure over
the solution decreases or
increases with increase in the
mole fraction of Component 1.

10. ● The composition of vapour phase in equilibrium with the solution is determined by the partial
pressures of the components.
● If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase
then,
● using Dalton’s law of partial pressures:
● P1 =y1 Ptotal
● p2 =y2 Ptotal
● In general Pi = yi Ptotal

11. Raoult’s law as a special case of Henry’s law
Henry’s law Raoult’s Law
p=KH x p1= p1
0
x1
In both the law we can see that partial pressure of the volatile
component is proportional to its mole fraction. Only the proportionality
constant is different. Thus Raoult’s law becomes a special case of
Henry’s law in which KH becomes equal to p1
0
.

13. Because of
difference in the
intermolecular
force of attraction
that arises when a
solute is added to a
solvent.
Why the graphs
obtained are not as
expected by Raoult

14. CASE I.
The A-B interactions are
weaker than A-A or B-B
interaction

15. In such a case molecules of A (or B) will find it easier to escape than in pure
state
This
will
➔ the vapour pressure of the solution is greater than
expected (as more molecules in vapour phase)
This type of deviation from Raoult’s law is termed as
POSITIVE DEVIATION
where the observed VP is greater than the calculated VP
( by Raoult’s law )

16. Mixtures of ethanol and acetone
In pure ethanol, molecules are hydrogen
bonded
On adding acetone, its molecules get in between
the ethanol molecules
hydrogen bonds in ethanol is broken.
Hence more molecules of (ethanol) will be there in the vapour phase
than the calculated from Raoult’s law.
The new bond between ethanol and acetone Will be
weaker than the original hydrogen bond

17. Examples and Graph
Examples:
❖ Acetone and CS2
❖ Acetone and Benzene
❖ Carbon tetrachloride and
Toluene or chloroform
❖ Methyl alcohol and water
❖ Acetone and ethanol
❖ Ethanol and water

18. ➔ ΔmixH > 0 because the heat absorbed will be more than the heat
released. (ΔmixH = +ve)
➔ ΔmixV > 0 because the volume expands on dissolution.
➔ They form MINIMUM BOILING AZEOTROPES

19. .
CASE II
In such a case molecules of
A (or B) will find it
DIFFICULT to escape than
in pure state.
This
will
The A-B interactions are stronger than A-A or B-B interaction
DECREASE the vapour
pressure than expected
(as less molecules in
vapour phase)

20. .
CASE II
➔ ΔmixH < 0 because more heat is released when new intermolecular
force of attraction are formed.
➔ ΔmixV < 0 because the volume decreases on dissolution.
This type of deviation from Raoult’s law is termed as
NEGATIVE
DEVIATION
where the observed VP is LESS than the calculated VP (by Raoult’s law)
They form MAXIMUM BOILING AZEOTROPES

21. Examples and Graph
➢ Chloroform and Benzene
➢ Chloroform and Diether
➢ Acetone and aniline
➢ Nitric acid and water
➢ Acetic acid and Pyridine
➢ HCl and water.

22. What is an ideal solution?
The solutions that obey’s Raoults law over the entire range of concentration
are known as Ideal solutions.
Characters:
ΔmixH = 0
ΔmixV = 0
A-A interactive force = B-B interacive force = A-B intereactive force
Examples:
1.Solutions of n-hexane and n-heptane 2. Bromoethane and chloroethane
3.Benzene and toluene

24. COMPARISION BETWEEN IDEAL AND REAL
SOLUTIONS

25. AZEOTROPES
Azeotropes are binary mixtures
having same composition in liquid
and vapour phase and boil at a
constant temperature.
In such cases it is not possible
to separate the components by
fractional distillation.
They are also referred as
boiling point mixtures.
Each azeotrope has a characteristic
boiling point. The boiling point of an
azeotrope is either less than the
boiling point temperatures of any of
its constituents or greater than the
boiling point of any of its
constituents .

26. TYPES OF AZEOTROPES
MINIMUM BOILING
AZEOTROPES
Ethanol boils at 78.4 °C, water boils at
100 °C, but the azeotrope boils at 78.2
°C, which is lower than either of its
constituents.[8]
Indeed, 78.2 °C is the
minimum temperature at which any
ethanol/water solution can boil at
atmospheric pressure
The solutions which show a large
positive deviation from Raoult’s law
form minimum boiling azeotropes
at a specific composition.
The boiling point of such an
azeotrope is less than the boiling
point temperatures of any of its
constituents.
95.63% ethanol and 4.37% water (by
mass) boils at 78.2 °C

27. MAXIMUM BOILING AZEOTROPES
The solutions which show a large
negative deviation from Raoult’s law
form maximum boiling azeotropes at
a specific composition.
The boiling point of such an
azeotrope is greater than the
boiling point of any of its
constituents
Example: hydrochloric acid
at a concentration of 20.2%
and 79.8% water (by mass).
HCl boils at −84 °C and water
at 100 °C, but the azeotrope
boils at 110 °C, which is
higher than either of its
constituents. The maximum
temperature at which any
hydrochloric acid solution can
boil is 110 °c

28. Vapour pressure of solutions of solids in
liquids:
In a solid solution the solute is solid which is non volatile whereas the solvent
is liquid which is volatile.
For such solutions some physical properties like vapour pressure, boiling point,
freezing point etc.will be different from those of pure solvents.
For instance,
The boiling point of water will increase if we add salt to it.
Similarly the freezing point of water also decreases on adding non volatile
solids to it.

29. Why does the physical properties
vary?

30. The number of solvent molecules on the surface will be now reduced, thus the
number of molecules going to the vapour phase will also be reduced. Consequently
p1< p1
0

31. ● The decrease in the VP of solvent depends on the QUANTITY ( No Of
Moles) of non volatile solute present in the solution rather than their
nature.
● The decrease in the VP of water by 1 mole of glucose is same as that
by 1 mole of sucrose , which will also be same by 1 mole of urea also.
● The decrease in the vapour pressure of water can be represented as
Δp , where Δp= p1
0
- p1
● Thus Δp1M glucose = Δp1M sucrose = Δp1M urea
● But no of moles of 1M urea and 1M NaCl will be different. Because

32. .
Raoult’s law in general format can be stated as “ For any solution the
partial vapour pressure of each volatile component in the solution
is directly proportional to its mole fraction”
P total = p1 +p2 P
⇒⇒⇒⇒ total = p1 = p1
0
x1 as p2 =0.
we
know
Thus the pressure of the solution depends only on the pressure of the solvent
as the solute is non volatile.
The vapour pressure of the solution at the given temperature is found to be
lower than the vapour pressure of the pure solvent. As p1<p1
0
This can be represented as Ptotal= Ps = p1 , where Ps is pressure of the
solution

33. Thus For a solid solution
P total = p1 = p1
0
x1
Ptotal = p1 = ps
● If a solution obeys Raoult’s law for all
concentrations, its vapour pressure would
vary linearly from zero to the vapour
pressure of the pure solvent.
● But when it comes to actual
solutions and the graph is plotted ,
what we get is different.

35. Ideal and Real SOlutions
Let us consider two components A and B.
In component A the inter molecular force of attraction between the
molecules can be represented as A-A type of interaction with a
particular strength.
Similarly, component B has B-B type of interaction.
When component A and B are mixed a new intermolecular attraction say A-B
is formed.

36. If the strength of A-B = (A-A) = (B-B) then
ΔmixH = 0 i.e enthalpy of mixing is zero. Means no energy as
heat is released or absorbed.
ΔmixV = 0 i.e volume of the solution will be equal to the sum of
volume of the component A and B.
When does this happen?
Such a solution is called IDEAL SOLUTION.
When A-A = B-B ( in terms of
intermolecular ENERGY)
Then we get the graph like this