1. Solutions are homogeneous mixtures of two or more components, where the component present in smaller amounts is called the solute and the primary liquid component is the solvent. 2. Solutes can be electrolytes, which dissociate into ions, or nonelectrolytes, which do not dissociate. Common methods to express the concentration of solutions include molarity, molality, mole fraction, and percent composition. 3. The solubility of solids in liquids and gases in liquids depends on factors like temperature, pressure, and the nature of the solute and solvent. Henry's law and Raoult's law describe gas solubility and vapor pressure lowering in solutions. Ideal solutions follow

1. 1
Unit:-2.Solutions:
:Laishram Dhanachandra singh ,Dept of Chemistry ,Vkv-Roing
Solutionsarehomogeneousmixtureoftwoormorethantwocomponents.Thosecomponentwhichtookinsmallest
quantityissolutewhiletheotherwhichhaslargecomponentarecalledsolvents.

2. Types ofsolutes
Na+
Cl-
• Electrolytes (Conductive):
Dissociation in to ions in
solution
Eg: NaCl
• Nonelectrolytes (no conductivity):
no dissociation
Eg: sugar

3. Concentration expressions for solutions
3
Mole fraction:
 Ppm= number of parts of the component in the solution x 106
Total number of parts of all the components of the solution
 Percent by weight % w/w= mass of the component in the solutionx100
Total mass of the solution
 Percent by volume %v/v= volume of the component in the solutionx100
Total volume of the solution
 Percent weight in volume % w/v=mass of the component in the
solutionx100
Total volume of the solution
Mole percent:-Mole percent = mole fraction X 100

4. Molarity:-It is defined as number of mole of solute dissolved in one litre of solution.
M= moles of solute/volume of solution in litre
Normality:-It is the number of equivalent of solute dissolved per litre of
solution. N= mass of solute/ equivalent mass of solute.
Eqivalent mass is the mass in gm that combines with displaces , directly or
indirectly,1.008g mass of hydrogen,8.00g of oxygen,35.5 gm of chlorine.
equivalent mass of acid= molar mass/basicity
equivalent mass of base= molar mass/acidity
equivalent mass of salt= molar mass/total +ve charge valency
equivalent mass of ions= molar mass/valency
equivalent mass of oxidising or reducing agent= molar mass/e-loss or gained
= molar mass/ change in oxidation number
equivalent mass of element= molar mass/valency
Normality equation=N1V1=N2V2
Molality:- It is defined as the number of moles of the solute per kg of the
solvent .m=moles of solute/mass of solvent in kg.
4
Molarity and normality both depend on the volume of the solvent, so their values
are affected by change of volume caused by factors such as change in
temperature. Molality doesn’t has this disadvantage.

5. 5
 Solubility: Maximum amount of solute dissolve in a specified amount of solvent
is called solubility. It depend on nature of solute , solvent, temperature and
pressure.
 Solubility of solid in liquid:- Solute dissolved in solvents which have similar inter
molecular interactions . The process which dissolved solute in solvent molecule
is called as dissolution. Solute particle in solution collide with some solute and
get separated . This process is called crystallization . In saturated solution , no
more solute can dissolve at that temperature and pressure. In unsaturated ,
some more solute can dissolved.
 Effect of temperature:-According to Le-Chateliers principle, any change in the
equilibrium brings changes which reduce the change and thereby bring another
dynamic equilibrium . In general dissociation process is
endothermic(ΔsolH>0)Solubility increase with rise in temperature and if it is
exothermic(ΔsolH>0)the solubility decrease.
 Effect of pressure:-It does not have any effect on solubility of solid in liquids . It is
so because solid and liquid are highly incompressible.

6.  Solubility of gas in a liquid: It depends on temperature and pressure. Solubility of gas
increase with increasing pressure.
 Henry’s Law:-The solubility of a gas in a liquid is directly proportional to the pressure
of the gas. The mole fraction of a gas in a solution is directly proportional to the partial
pressure of the gas over the solution. The partial pressure of the gas in vapour
phase(p) is proportional to the mole fraction of the gas (X) in the solution. P=KHX
 Higher the value of KH at a given pressure , the lower is the solubility of the gas in
the liquid. The solubility of the gas in the liquid decreases with increase in
temperature.
 Application of Henry’s Law:-1.To increase the solubility of CO2 in soft drink
and soda water the bottle sealed under high pressure.
 2. Scuba driver experience high pressure under water. Increased pressure
increases solubility of air in blood . When driver come on surface , pressure
decrease .This releases the dissolved gases and leads to formation of
bubbles of nitrogen. This blocks capillaries and creates a condition known as
bends which results painful. To avoids this toxic effects the tanks filled with
air diluted with helium.
 3.At high altitudes the partial pressure of oxygen is less than ground level which leads
to weak and unable to think clearly ,symptoms of a condition known as anoxia

7. Effect of temperature: Solubility of gas in liquids decreases with rise in
temperature, when dissolved ,the gas molecules are present in liquid phase
and the process of dissolution ,heat is evolved.
7
 Vapour phase of liquid solution:- Raoult’s law state that for a solution of volatile liquids the
partial pressure of each component in the solution is directly proportional to its mole
fraction . For component 1,P1=x1 and p1=P0
1x1 Where P0
1 is the vapour pressure of pure
component 1 at the same temperature.
Similarly for component 2
P2=P0
2x2 Where P0
2 is the vapour pressure of pure component 2 at the same temperature.
According to Dalton law of partial pressures, the total pressure(Ptotal) over the solution
phase in the container will be the sum of the partial pressure of the component of the
solution and is given as(Ptotal)=p1+p2
Substituting the value of p1 and p2, we get
(Ptotal)= P0
1x1+ P0
2x2
=(1-x2) P0
1 + x2 P0
2 = P0
1 +( P0
2- P0
1) x2
minimum value of P -total is at P0
1
maximum value of (Ptotal) is at P0
2 P0
2
If 1 is less volatile than 2 , P0
1 < P0
2 P0
1
The composition of vapour phase in equilibriumwith the solution is
determine by partial pressures of the component s.
P1= y1 (Ptotal) vapour pressure
P2= y2 (Ptotal) X1=1 X2 X1=0
 in general X2=0 X2=1
 Pi= yi (Ptotal)
Ptotal =P1+P2(III)
P1(I) P2(II)

8. partial pressure of each component in the solution is directly proportional to its mole fraction .In the solution of a gas in liquid,one
of the component is so volatile that it exist as a gas and the solubility is givenby Henry,s law as partial pressure of the gas in
vapour phase(p) is proportional to the mole fraction of the gas (X) in the solution. P=KHX
 By comparing these two law we get the partial presure of the volatile component is directly propertional to mole
fraction in solution.Only proportional constant. KH differ from P0
1
 Raoult’s law become a special case of Henry,s law in which KH become equal to P0
1 .
 Vapour pressure of solution of solid in liquids:- A liquid vapourised at a given temperature and under
equilibrium conditions the pressure exerted by the vapours of the liquid over the liquid phase is called vapour
pressure. In a pure liquid the entire surface is occupied by the molecules of the liquid. If non volatile solid is
added to a solvent to give a solution ,the vapour pressure is only from solvent, thus the vapour pressure is
lower than that of pure solvent as in the solution surface both the solute and solvent are present .Hence no of
solvent molecule escape from the surface reduced and the vapour pressure reduced.
 Thus the vapour pressure depends on amount of non valatile solute present in solutions. Thus in general
Raoult’s law state as For any solution the partial vapour pressure of each volatile component in the solution is
directly proportional to its mole fraction.
 According to Raoult’s law P1∞X1
 And Pi=Xi P0
1 The proportionality constant is equal to the vapour pressure of pure solvent and the plot of vp
vs mole fraction is linear.
 vapourpressure
 0 X solvent 1

9. Ideal and real solutions
9
 Ideal solution is defined as a solution in which there is no change in
the properties of the components other than dilution when they are
mixed to form the solution . The solution which obey Roult’s law over
the entire range of concentration are known as ideal solution . An
ideal solution has two propertie(i) enthalpy of mixing of the pure
component to form the solution is zero and (ii) the volume of mixing
is also zero.
 ∆mixH=0, ∆mixV=0 .
 If A-A interaction &B-B interaction = A-B interaction then the solution
shows ideal character but it is very rear. e,g: solution of n-hexane
and
n–heptane, bromo ethane and chloro ethane, benzene and toluene.
 Molecules exhibit complete freedom of motion and randomness
of distribution in the solution.
 Ideality in solutions means complete uniformity of attractive forces.

10. Ideal Solutions and
10
 P = pA + pB
 pA = pA◦ XA
 pB = pB◦ XB
 What is the partial vapor pressure of benzene and of
ethylene chloride in a solution at a mole fraction of
benzene of 0.6? The vapor pressure of pure benzene at
50◦C is 268 mm, and the corresponding pA ◦ for
ethylene chloride is 236 mm.
Raoult's Law States that, in an ideal solution, the partial vapor pressure of each volatile
constituent is equal to the vapor pressure of the pure constituent multiplied by its
mole fraction in the solution. Thus, for two constituents A and B,

11. 11
Vapor pressure composition curve (for
previous example)

12. Real Solutions
12
 Ideality in solutions presupposes complete uniformity
of attractive forces.
 Many examples of solution pairs are known,
however, in which the “cohesive” attraction of A for A
exceeds the “adhesive” attraction existing between A
and B.
 Similarly, the attractive forces between A and B may
be greater than those between A and A or B and B.
 Such mixtures are real or nonideal; that is, they do
not adhere to Raoult’s law
 Two types of deviation from Raoult’s law are
recognized, negative deviation and positive

13. 13
When the “adhesive”
attractions between
molecules of different
species exceed the
“cohesive” attractions
between like molecules,
the vapor pressure of
the solution is less than
that expected from
Raoult’s ideal solution
law, and negative
deviation occurs.
Negative deviation
Adhesion > Cohesion
A-B weaker than A-A&B-B
Eg. Ethanol and Acetone, acetone
&CS2
Ethanol has H bonding addition
of acetone break this and weaken
the force and thus shows +ve
deviations.

14. 14
When the “adhesive”
attractions between
molecules of different
species are weaker
than “cohesive”
attractions between like
molecules, the vapor
pressure of the solution
is more than that
expected from Raoult’s
ideal solution law, and
positive deviation
occurs.
Positive deviation
Adhesion < Cohesion i,e A-B greater than
A-A&B-B leading to decrease in v.p
resulting –ve deviation.
E,g : phenol and aniline
Here H-bonding between proton and
nitrogen in aniline is stronger than
aniline.Chloroform and acetone as
Chlorofom can form H-bond with acetone.

15. .Liquid mixture having the same composition in liquid and vapour phase and boil at
constant temperature are called azeotropes mixture.
.Solutions which shows a large +ve deviations from minimum boiling azeotropes at a
specific composition.eg ethanol and
water (95% vol).
.Solutions which shows a large -ve deviations from miximum boiling azeotropes at a
specific composition.eg ;6 8% nitric acid and 32% water with a boiling point of 393.5K .
Colligative properties
15

16. Colligative properties
16
 Colligative properties of solutions are those that affected (changed) by the presence of solute and
depend solely on the number (amount of solute in the solutions) rather than nature of constituents.
Examples of colligative properties are:
1. Relative lowering of Vapor pressure
2.Elevation of Boiling point
3. Depression of Freezing point
4.Osmotic pressure
 1. Relative lowering of Vapor pressure:- 509393 A
 Lowering of vapour pressure depends on concentration of solute .The relation between v.p of solution
,mole fraction and V.P of solvent as P1=X1P0
1
The reduction in the v.p of solvent (∆P1 ) is given as: ∆P1 = P0
1 - P1 = P0
1 (1- x1 )……(equation 1)
Knowing that X2=1-X1 )(equation 1)reduces to ∆P1 = x2 P0
1 …… (equation I1)
In a solution containing several non volatile solutes,the lowering of the vapour pressure depends on the
sum of the mole fraction of different solutes. (equation I1)can be written as ∆P1 = P0
1 - P1 = x2
…..(equation I1I)
P0
1 P0
1
The expression P0
1 - P1 is called relative lowering of vapour presssure. And is equal to mole
fraction.
P0
1
 P0
1 - P1 = n2 = W2x M1
 P0
1 n2+n1 W1XM2

17. Colligative vs Non-colligative
17
Compare 1.0 M aqueous sugar solution to a 0.5 M solution of
salt (NaCl) in water.
both solutions have the same number of dissolved particles
any difference in the properties of those two solutions is due to
a non-colligative property.
Both have the same freezing point, boiling point, vapor
pressure, and osmotic pressure

18. Non-Colligative Properties
Sugar solution is sweet and salt solution is salty.
Therefore, the taste of the solution is not a colligative
property.
Another non-colligative property is the color of a solution.
Other non-colligative properties include viscosity, surface
tension, and solubility.
18

19. Lowering of vapor pressure
Vapor pressure:
Pure solvent > solutions

20. Lowering of vapor pressure
20
 According to raoult’s law Psolvent = Pºsolvent Xsolvent
 But if the solute used in non volatile only pressure from
solvent can be considered.
 So:
 On the other hand
Psolute = Pºsolute Xsolute
Psolution = Pºsolvent Xsolvent
X1 = mole fraction of solvent
X2 = mole fraction of solute

21. 21
 ∆p = p1◦ − p is the lowering of the vapor pressure
and ∆p/p1◦ is the relative vapor pressure lowering.
 The relative vapor pressure lowering depends only
on the mole fraction of the solute, X2, that is, on the
number of solute particles in a definite volume of
solution. Therefore, the relative vapor pressure
lowering is a colligative property.

22. 22
 Calculate the relative vapor pressure lowering at 20◦C
for a solution containing 171.2 g of sucrose (w2) in 100
g (w1) of water. The molecular weight of sucrose (M2)
is 342.3 and the molecular weight of water (M1) is
18.02 g/mole.

23. Boiling point elevation
23
Boiling point elevation is a colligative property related to
vapor pressure lowering.
The boiling point is defined as the temperature at which
the vapor pressure of a liquid equals the atmospheric
pressure.
Due to vapor pressure lowering, a solution will require a
higher temperature to reach its boiling point than the pure
solvent.

24. Elevation of the Boiling Point
24
 The boiling point of a solution of a nonvolatile solute
is higher than that of the pure solvent owing to the
fact that the solute lowers the vapor pressure of the
solvent.
ΔTb = K X2
ΔTb = Kbm
boiling point is a colligative property

25. 25
 In dilute solutions: The elevation of b.pt ∞ molal concentration of the solute
in a solutionThus = ∆ Tb ∞ m or ∆ Tb = Kb m .the unit of Kb is k kg mol-1
m = w2 /M2 = 1000 X W2
w1/1000 M2 x W1
Substituting the value of molality in equation ∆ Tb = Kb m
∆ Tb = Kb X 1000 xW2
M2 X W1
Thus M2 = 1000 X W2 X Kb
 ∆ Tb X W1
ΔTb = K X2 ΔTb = Kbm
Tb: is known as the boiling point elevation
Kb: is called the molal elevation constant.
m: is molality of solvent

26. Freezing Point
26
Every liquid has a freezing point - the temperature at
which a liquid undergoes a phase change from liquid to
solid.
When solutes are added to a liquid, forming a solution, the
solute molecules disrupt the formation of crystals of the
solvent.
That disruption in the freezing process results in a
depression of the freezing point for the solution relative to
the pure solvent.

27. Depression of the Freezing Point:-Decrease in freezing point due to addition of non volatile
solute is called depression in freezing point ∆T f = Tº f – T f
27
For a dilute solution(ideal solution) depression in freezing point is directly proportional to molality.This
∆T f ∞ m = ∆Tº f= K f m .The unit of Kf is K Kg mol-1.If w2 gm solutehaving M2 mass is present
in W1 gm of solvent produces depression in freezing point ∆T f of the solvent then molalityis given
by formula m= W2/M2
W1/1000
Substituting this value we get ∆T f = Kf X W2/M2
W1x1000
∆T f = Kf X W2X 1000
M2X W1
M2= Kf X W2X 1000 Kf and Kb depends on nature of solvent
∆T f X W1
Kf =R X M1 X Tf 2 , Kb =R X M1 X Tb 2
1000 xΔΔH fus 1000 xΔΔH vap
Kf is the molal epression constant

28. 28
What happens to the triple point?

29. 29

30. Osmotic Pressure
30
When a solution is separated
from a volume of pure solvent
by a semi-permeable
membrane that allows only the
passage of solvent molecules,
the height of the solution
begins to rise.
The value of the height
difference between the two
compartments reflects a
property called the osmotic
pressure of a solution.

31. Osmotic Pressure
Where
π is the osmotic pressure .
V is the volume of the solution in liters.
n is the number of moles of solute.
R is the gas constant, equal to 0.082 liter atm/mole deg.
T is the absolute temperature.
 Van't Hoff and Morse Equations for Osmotic Pressure:Osmotic pressure is proportional to the molarity,c
of the solution at a given temperature ,Thus
π=CRT Since C=n/V thus, π=(n2/v)RT
But, v volume of solution in litre contain n2 moles of solute if w2 gm of solute of molar
masssbM2 is present in the solution .Then n2= W2/M2,thus , πV= W2RT
M2
Hence M2= W2RT
 πV
Same osmotic pressure –isotonic (o.p of blood solution has exactly equal with 0.9% NaCl
solution) thus it is normal saline solution , less salt concentration is hypotonic(less than
0.9%) , while high salt concentration is Hypertonic(more than 0.9%)

32. Abnormal molar masses:
During association or dissociation the molar mass is either higher or lower ,this is called
Abnormal molar mass. Van’t Hoff factor is introduce to account association or
dissociation and indicated by I
i= Normal molar mass/Abnormal molar mass
= observed colligative property/ calculated colligative property
= total no.of moles os particle after association or dissociation
No.of moles of particle after association or dissociation
If i= less than1,association take place, If i= greater than1,dissociation take place
P0
1 - P1 = i n2
P0
1 n1
Elevation of b.pt= ∆ Tb =i Kb m
Depression of freezing point= ∆ Tf =i Kf m
Osmotic pressure= , π=in2RT/V
32

33. MOLECULAR WEIGHT DETERMINATION
33
 The four colligative properties can be used to calculate
the molecular weights of nonelectrolytes present as
solutes. Using vapor pressure lowering
Using boining point elevation

34. 34
Using Freezing point depression
M2 =
𝑔 𝑅𝑇
Π
Using Osmotic pressure

35. 35

36. 36

37. 37