SOLUTIONS

Solution.

A solution is a homogeneous mixture of two or more substances whose composition can be varied
within a certain limit.

Solvent. The component present in excess and whose physical state is same as that of solution is called
a solvent.

Solute. The component present in the smaller amount is called as solute.

Characteristics of solution.

1. A solution is homogeneous.
2. Particles of solute cannot be seen by the naked eye.
3. Solution does not allow the beam of light to scatter.[does not show Tyndall effect]
4. A solution is stable.
5. The solute from a solution cannot be separated by a filtration.
6. Solution is composed of only one phase.

Types of solutions.

1. Sold-Liquid solutions. In such solutions, solid is referred to as solute and liquid as solvent.
Examples 1. Sugar dissolved in water.
2. Salt dissolved in water.
2. Liquid – Liquid solutions. In such solutions, both solute and solvent are liquids and are
chemically alike.
A polar liquid is completely miscible with some other polar liquid.
Example: Ethanol in water.
A non-polar liquid is soluble in some other non-polar liquid.
Example: benzene in toluene.
3. Gas -Liquid solutions. In such solutions, gas is referred to as solute and liquid as solvent.
Solubility of a gas in a liquid increases with increase in pressure and decreases with increase in
temperature.
4. Solid solutions. Homogeneous mixture of two or more solids is termed as solid solutions

There are two types of solid solutions.

A) Substitutional solid solutions. In this type, constituent particles present in the crystal
lattice of one substance are replaced by similar particles of some other substance.
Examples: Steel, brass, bronze etc
B) Interstitial solid solutions. In this type atoms of one substance are placed into the voids
or interstitial sites of some other substance. Example: tungsten carbide (WC)

Methods of Expressing The Concentration of Solutions.

The amount of solute present in a given quantity of solution is called concentration.

It is expressed in several ways.

1. Molarity (M).
The number of moles of solute dissolved per litre of a solution is called molarity.

Molarity changes with temperature

Unit of molarity is mol L-1

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑛𝐵
Molarity = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒 OR M= 𝑉

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔 𝑤𝐵
M = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒 OR M= 𝑀
𝐵𝑥𝑉

𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑖𝑛 𝑔𝑟𝑎𝑚 𝑝𝑒𝑟 𝑙𝑖𝑡𝑟𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Molarity = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

Mass of solute = M x MB x V

2. Molality(m).
The number of moles of solute dissolved per kilogram or 1000 gram of solvent is called molality.

Molality does not change with temperature.

Unit of molality is mol kg-1

𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝑩 𝒏
Molality = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒊𝒏 𝒌𝒈
OR m = 𝑾𝑨

𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒘𝑩
Molality= 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒙 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒊𝒏 𝒌𝒈
OR m = 𝑀𝐵 𝒙𝑾𝑨

𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒙 𝟏𝟎𝟎𝟎 𝒘𝑩 𝒙 𝟏𝟎𝟎𝟎

Molality = OR m =
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒙 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒊𝒏 𝒈 𝑀𝐵 𝒙𝒘𝑨

3. Normality(N).
The number of gram equivalents of solute dissolved per litre of a solution is called normality.
Unit of normality is gram eq. L-1
Normality changes with temperature.
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒈𝒓𝒂𝒎 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝑛𝑒𝑞
Normality (N) = OR N = =
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒍𝒊𝒕𝒓𝒆𝒔 𝑉
𝒘𝑩 𝒙 𝟏𝟎𝟎𝟎
OR N = 𝑬𝒙𝒗

𝑵𝒙𝑬𝒙𝒗
Mass of solute= N x E x V = 𝟏𝟎𝟎𝟎

Relation between Normality and Molarity

Normality x Equivalent mass = Molarity x Molecular mass

𝑵𝒐𝒓𝒎𝒂𝒍𝒊𝒕𝒚 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔

𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚
= 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒎𝒂𝒔𝒔 = 𝒏𝒇𝒂𝒄𝒕𝒐𝒓
𝒏𝒇𝒂𝒄𝒕𝒐𝒓 = basicity for acids
= Acidity for bases
= Total charge of the cationic or anionic part for salts.
 4. Mole fraction (Χ). Mole fraction of a particular component is the ratio of number of moles of
that component to the total number of moles of all the components present in the solution.
Mole fraction is independent of temperature.

𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒕

Mole fraction of a component = 𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒂𝒍𝒍 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔

For a binary solution containing components A and B:

𝑛𝐴
Mole fraction of A, XA = 𝑛 +𝑛
𝐴 𝐵

𝑛𝐵
Mole fraction of B, XB =
𝑛𝐴 +𝑛𝐵
Sum of mole fractions of all the components in a solution is always unity.
XA + XB = 1
Ratio of mole fractions of any two components is equal to the ratio of their number of moles.
𝑿𝑨 𝑛𝐴
=
𝑿𝑩 𝑛𝐵
5. Parts per Million (ppm).
It is the mass of solute in million parts by mass of the solution.
ppm of a solute = mass of solute x 106
Mass of solution.

Note:

• Molality is preferred to molarity to express the strength of solutions. This is because, molality
involves the masses of solute and solvent whereas molarity involves volume of solutions.
Temperature has no effect on mass but volume of solution can change with temperature.
Hence, molality, mole fraction and mass fraction do not change with temperature whereas
molarity, normality and formality change with temperature.
• 1 molar aqueous solution has higher concentration than 1 molal solution. A 1 molar solution
contains one mole of solute in one litre (1000mL) of solution while one molal solution contains
one mole of solute in 1000 g of solvent. If density of water is 1 then one mole of solute is
present in 1000 mL of water in 1 molal solution while one mole of solute is present in less than
1000mL of water in 1 molar solution. (1000 mL solution = amount of solute + amount of
solvent.) Hence number of moles of solute per mL will be less in 1 molal solution than in 1 molar
solution. Thus 1 molar solution is more concentrated than 1molal solution.

Solubility of Gases in Liquids.

Solubility of gases in liquids depends on:

a) Nature of the gas and solvent.
Generally, gases which can be easily liquified, are more soluble in polar solvents
like water.
For example, CO2 is more soluble in water than O2, N2 or H2.
Gases which react with solvent possess high solubility.
For example, NH3, HCl, CO2, SO2 etc are highly soluble in water.
 b) Temperature.
Solubility of gases generally decreases with increase in temperature.
c) Pressure.
Solubility of gases increases with increase in pressure.

Henry’s law.

1. Solubility of a gas in a liquid is directly proportional to the pressure of the gas in equilibrium
with the solution
Solubility ∝ Pressure of the gas.

If solubility is expressed in mole fraction of gas:

2. Mole fraction of the gas in a solution is directly proportional to the partial pressure of the gas
over the solution.
OR
3. Partial pressure of a gas in vapour phase is directly proportional to the mole fraction of the
gas in the solution.
P∝X

P = KH X

KH = Henry’s law constant

Significance of KH

1. Value of KH depends on the nature of the gas.

2. Higher the value of KH, lower is the solubility of the gas in the liquid.
𝟏
KH ∝ 𝑿
3. The value of KH increases with increase in temperature.

4. The solubility of most of the gases (except H2 and inert gases) in a liquid decrease with increase
in temperature and therefore, Henry's law constants have smaller values at higher
temperatures.

5. Aquatic species like fish feel more comfortable in lakes in winter than in the summer.
Aquatic species require dissolved O2 for breathing. As solubility of gases decreases with increase
of temperature, less oxygen is available in summer than in winter. Hence, they feel more
comfortable in winter (low temperature) when the solubility of O2 is higher.

Graphical representation of Henry’s Law.

 Applications of Henry’s Law

1. In the production of carbonated beverages or soft drinks.

According to Henry’s law solubility of a gas in a liquid is directly proportional to the pressure
of the gas. Hence, to increase the solubility of carbon dioxide in soft drinks, the bottles are
sealed under high pressure. When bottle is opened, the pressure inside falls to atmospheric
pressure and the excess carbon dioxide bubbles out of the solution causing effervescence.

2. In deep sea diving or Scuba diving.

If deep sea divers use air for respiration, they develop a medical condition known as bends.
[Compressed air contains nitrogen in addition to oxygen. According to Henry’s law, solubility of
gases increases with the increase in pressure. At great depths when the diver breathes the
compressed air, more nitrogen dissolves in the blood because of high pressure. When the diver
comes towards the surface, the pressure decreases and nitrogen comes out, forms bubbles in
the blood. These bubbles restrict blood flow and can even burst the blood vessels. This is quite
painful and dangerous to life. This condition is known as bends.] In order to avoid bends, the
tank used by sea divers are filled with air diluted with helium because of the lower solubility
of helium in blood than nitrogen.
3. At high altitudes, the partial pressure of oxygen is less than that at the ground level. This results
in low concentration of oxygen in the blood of people living at high altitudes. The low
concentration of oxygen in blood leads to weakness and loss in the clarity of thinking known
as anoxia.

Limitations of Henry’s Law

1. It is applicable to ideal gases only. (It is applicable to real gases at low pressure because, real
gases approach ideal behavior at low pressure).
2. It is not applicable to gases which undergo chemical change in solution. Example: ammonia.
3. The law cannot be applied to those gases which dissociate into ions. Example: HCl.

Vapor pressure

The pressure exerted by the vapours in equilibrium with its liquid at a given temperature is known as
vapor pressure.

Vapour pressure of a liquid depends on:

1. Nature of the liquid or inter molecular attractive force.

The liquids which have weaker inter-molecular attractive forces possess greater vapour
pressure.
2. Temperature.
The vapour pressure of a liquid increases with increase in temperature.

[I] Raoult’s Law for solution containing volatile liquids (volatile solute).
Statement 1.
For a solution of volatile liquids, partial vapor pressure of each component in a solution is directly
proportional to the mole fraction of the component in the solution.
For example, for a binary solution containing volatile components A and B,

Partial pressure of A, PA ∝ XA

PA = constant x XA

If XA = 1, ie the component A is pure state, PA = 𝑃𝐴0

Ie 𝑃𝐴0 = constant. 1

Hence, the value of constant = PoA, the vapour pressure pure component A

Thus, Partial pressure of component A.

PA = 𝑷𝟎𝑨 x XA

Similarly, Partial pressure of B, PB ∝ XB

PB = 𝑷𝟎𝑩 x XB

Statement 2.

For a solution containing volatile liquids, the partial vapour pressure of each component in the
solution at a particular temperature is equal to the product of vapour pressure of pure component
and its mole fraction in the solution.

PA = 𝑃𝐴0 x XA

PB = 𝑃𝐵0 x XB

Total vapour Pressure of the solution, P = PA + PB = 𝑃𝐴0 x XA + 𝑃𝐵0 x XB ----------------- 3

P = PA + PB = 𝑃𝐴0 x XA + 𝑃𝐵0 x XB

Effect of addition of non-volatile solute to a solvent.

When a non-volatile solute is added to a volatile liquid (solvent), the vapour pressure of the liquid
gets lowered.
Or

The vapour pressure of a solution containing non-volatile solute is always lower than that of the pure
solvent.

Explanation:

In the case of a pure solvent, the entire surface of the liquid is occupied by the solvent molecules. But in
the case of a solution containing non-volatile solute, a part of the liquid surface is also occupied by the
solute particles. As a result, the area of the surface containing liquid molecules decreases. This reduces
the escaping tendency of the solvent molecules in the solution. Thus, evaporation of solvent in a
solution takes place to a lesser extent. This results in the lowering of vapour pressure of the solution.

The difference between the vapour pressure of pure solvent and that of solution is called lowering of
vapour pressure. ( 𝑷𝟎𝑨 – P) = ΔP

The ratio of lowering of vapour pressure to the vapour pressure of pure solvent is called relative
𝑷𝒐𝑨 −𝐏
lowering of vapour pressure. ( 𝑷𝒐𝑨
)

[II] Raoult’s Law for solution containing non-volatile solute.

We know that, for a solution containing volatile liquids,

vapour pressure of solution, P = PA + PB

Or P = 𝑷𝟎𝑨 x XA + 𝑷𝟎𝑩 x XB

If the solute is non-volatile, 𝑷𝟎𝑩 = 0

Therefore, PB = 𝑃𝐵0 x XB = 0

Thus, vopour pressure of the solution,

P = 𝑷𝟎𝑨 . XA

Statement 1

The vapour pressure of a solution containing non-volatile solute at a particular temperature is equal
to the product of the vapour pressure of solvent in pure state and its mole fraction in the solution.

P = 𝑷𝟎𝑨 . XA

Or P ∝ XA

Statement 2.

The vapour pressure of a solution containing non-volatile solute at a particular temperature is directly
proportional to the mole fraction of solvent in the solution.

P = 𝑷𝟎𝑨 . XA
𝑷
𝑷𝟎𝑨
= XA But XA = (1-XB)

𝑷 𝑷
Therefore 𝑷𝟎𝑨
= (1-XB) Or 1- 𝑷𝟎𝑨
= XB

Or Relative lowering of vapour pressure,

𝑷𝒐𝑨 −𝐏
(𝑷𝒐𝑨
) = XB
Statement 3.

The relative lowering of vapour pressure of a solution containing non-volatile solute is equal to the
𝑷𝒐𝑨 −𝐏
mole fraction of the solute in the solution. ( ) = XB
𝑷𝒐𝑨

Ideal Solutions.

Solutions which obey Raoult's law at all concentrations and at all temperatures are called ideal
solutions.

Ideal solutions are formed by mixing two components which are

1) identical in molecular size

2) identical in molecular structure
3) identical in polarity
4) of almost identical intermolecular forces.

Two liquids A and B can form ideal solutions only when the A…..B intermolecular forces of attractions
are similar to A….A and B…..B intermolecular forces.

A perfect ideal solution is rare but several solutions exhibit nearly ideal behavior.

Examples:

1. benzene + toluene 3) chlorobenzene + bromobenzene

2. n- hexane + n- heptane 4) ethyl bromide + ethyl iodide

Characteristics of ideal solutions

1. Solution and its components obey Raoult’s law.

PA = 𝑃𝐴0 x XA PB = 𝑃𝐵0 x XB P = 𝑷𝟎𝑨 x XA + 𝑷𝟎𝑩 x XB

2. Change in volume on mixing is zero. ∆Vmixing = 0

ie, the total volume of the solution is equal to the sum of the volumes of its individual
components.
3. Change in enthalpy on mixing is zero. ∆Hmixing = 0.
ie, no heat is absorbed or involved when components are mixed.

Vapour pressure – Composition diagram for Ideal Solutions.

• The graph obtained by plotting vapor pressure against to mole fraction is a straight line. This
graph is called p-x diagram
• Dotted lines RQ and PS give the plot of partial vapour pressure against mole fraction.
• The solid line RS represents the plot of total vapour pressure against mole fraction
• In the case of ideal solution, the vapour pressure of solution at different composition lies
between the vapour pressures of pure components.
• P-X diagram is helpful in determining the values of PA , PB and P for the solutions of a particular
composition.

PA = 𝑃𝐴0 x XA

PB = 𝑃𝐵0 x XB

P = 𝑷𝟎𝑨 x XA + 𝑷𝟎𝑩 x XB

• The components of an ideal solution can be separated by fractional distillation

• When fractionally distilled, less volatile component (A) is obtained as residue and the more
volatile component(B) is obtained as distillate.

Note.

Separation of liquids by fractional distillation is possible only when the vapour phase has a
composition different from that of liquid phase (boiling liquid mixture).

Distillate: Liquid obtained by cooling the vapours

Residue: Liquid left in the flask.

Non-ideal Solutions.
Solutions which do not obey Raoult’s law at all concentrations and temperatures are called
non-ideal solutions.

Characteristics of non-ideal solutions.

1. They do not obey Raoult’s law.

PA ≠ 𝑃𝐴0 x XA

PB ≠ 𝑃𝐵0 x XB

P ≠ 𝑷𝟎𝑨 x XA + 𝑷𝟎𝑩 x XB

2. The change in volume on mixing is not equal to zero.

∆Vmixing ≠ 0
3. Change in enthalpy on mixing is not equal to zero
∆Hmixing ≠ 0
Types of non-ideal solutions.

Type-I – Non-ideal solutions showing positive deviation from Raoult’s law.

P-X diagram

• Vapour pressure curves of individual components and solution lie above the ideal solution
curve.
• The total vapor pressure of the solution rises to a maximum.
• At this point, vapor pressure is higher than those of pure components. (𝑃𝐴𝑜 &𝑃𝐵𝑜 )
• For any mole fraction the total vapour pressure (also partial vapor pressure) is more than the
value expected for an ideal solution of same composition.
PA > 𝑃𝐴0 x XA PB > 𝑃𝐵0 x XB P > 𝑃𝐴0 x XA + 𝑃𝐵0 x XB
• ∆Hmixing > 0 (+ve)
Dissolution of components A and B is accompanied by absorption of heat energy (endothermic).
• ∆Vmixing > 0 (+ve)
Volume of solution is greater than the the total volume of components before mixing.
• Examples
water + ethanol
CCl4 + CHCl3
Water + CS2
Acetone+ ethanol
• Boiling points such mixtures will be less than that of ideal solution of same composition.
• This type of solutions from minimum boiling azeotropes.
• On fractional distillation only one component is obtained in the pure state and the other
fraction will be azeotropic mixture.

P-X diagram. The graph obtained by plotting vapour pressure of solution(P) and the mole fraction
of(X) of components in the liquid phase and vapour phase (at constant temperature) is called P-X
diagram.(vapour pressure-composition diagram)
T-X diagram.

The graph obtained by plotting boiling point of solution(T) and the mole fraction of(X) of components
in the liquid phase and vapour phase (at constant pressure) is called T-X diagram. (temperature-
composition diagram)

P-X diagram. T-X diagram.

• A solution of composition between A and M on fractional distillation will give a residue of pure
A and a distillate of composition M.
• Similarly, a solution of composition between B and M on fractional distillation will give a residue
of pure B and a distillate of azeotropic mixture of composition M.

Explanation for Positive Deviation.

• This type of deviation is shown by liquid pairs for which the A....B molecular interaction forces
are weaker than A …A or B…..B molecular interaction forces.
• When the AB molecular interaction forces are weaker, the molecules of A or B find it easier to
escape from solution than from the components. This results in an increase in the vapour
pressure of the solution.
• Since A.....B molecular forces are weaker, the molecules will be loosely held and therefore,
there will be an increase in the volume during the formation of solution. This is why ∆Vmixing is
positive in this case.
• Since the A…B molecular interaction forces are weaker than A....A or B.....B molecular
interaction forces, more energy is required to overcome A ….A or B….B interaction forces. This
energy is more than that released in the formation of A...B interaction forces. Hence the
dissolution is endothermic and ∆Hmixing is positive in this case.

AZEOTROPIC MIXTURE.

A Solution of completely miscible liquids which boils at a constant temperature and distils over without
any change in composition is called an azeotropic mixture.
Minimum boiling Azeotropic mixture.

• This type of azeotropic mixture is formed by non-ideal solutions which show positive deviation
from Raoult’s law.
• In this type, the vapour pressure -composition curve has a maximum (M).
• A solution having higher vapour pressure boils at a lower temperature.
• Therefore, boiling point of the solution is at the point M would be minimum and even less than
the boiling points of pure components.
• At this point, solution of composition M has the same composition as that of vapour phase and
therefore would distil over without any change in composition.
• Thus, the solution corresponding to the composition M is called minimum boiling Azeotropic
mixture.
• For example, a mixture of ethyl alcohol and water containing 95.57%(95.6%) ethanol is a
minimum boiling Azeotropic mixture. This mixture boils at 351.15K which is less than the boiling
points of water (373K) and ethanol (351.3K). The distillate obtained on fractional distillation
contains same percentage of ethyl alcohol (95.6%). Thus, ethyl alcohol and water cannot be
separated from the mixture by fractional distillation.

Type -II – Non-ideal solutions showing negative deviation from Raoult’s law.

P-X diagram

• Vapour pressure curves of individual components and solution lie below the ideal solution curve.
• The total vapour pressure of the solution falls to a minimum.
• At this point, vapour pressure is lower than those of pure components. (𝑃𝐴𝑜 &𝑃𝐵𝑜 )
• For any mole fraction the total vapour pressure (also partial vapor pressure) is less than the
value expected for an ideal solution.

PA < 𝑃𝐴0 x XA PB < 𝑃𝐵0 x XB P < 𝑃𝐴0 x XA + 𝑃𝐵0 x XB

• ∆Hmixing < 0 (-ve)

Dissolution of components A and B is accompanied by absorption of heat energy
(endothermic).
 • ∆Vmixing < 0 (-ve). Volume of solution is greater than the total volume of components before
mixing.
• Examples
1) Water + HCl 2) C6H6 + CHCl3 3) Water + HNO3
• Boiling points such mixtures will be higher than that of ideal solution of same composition.
• This type of solutions from maximum boiling azeotropes.
• On fractional distillation only one component is obtained in the pure state and the other fraction
will be azeotropic mixture.
P-X diagram. T-X diagram.

• A solution of composition between A and M on fractional distillation will give pure A as distillate
and azeotropic mixture of composition M as residue.
• Similarly, a solution of composition between B and M on fractional distillation gives pure B as
distillate and azeotropic mixture of composition M as residue.

Explanation for Positive Deviation.

• This type of deviation is shown by liquid pairs for which the A….B molecular interaction forces are
stronger than A …A or B…..B molecular interaction forces.
• When the A….B molecular interaction forces are stronger, the molecules of A or B find it difficult
to escape from solution than from the components. This results in a decrease in the vapour
pressure of the solution.
• Since A…..B molecular forces are stronger, the molecules will be tightly held and therefore there
will be a decrease in the volume during the formation of solution. This is why ∆Vmixing is
negative in this case.
• Since the A…B molecular interaction forces are stronger than A….A or B…..B molecular
interaction forces, more energy is released during the formation of A …...B interaction forces.
This energy is more than that absorbed in the breaking of A…B interaction forces. Hence the
dissolution is exothermic and ∆Hmixing is negative in this case.
Minimum boiling Azeotropic mixture.

• This type of azeotropic mixture is formed by non-ideal solutions which show negative deviation
from Raoult’s law.
• In this type, the vapour pressure -composition curve has a minimum (M).
• A solution having lower vapour pressure boils at a higher temperature.
• Therefore, boiling point of the solution at the point M would be maximum and even more than
the boiling points of pure components.
• At this point, solution of composition M has the same composition as that of vapour phase and
therefore would distil over without any change in composition.
• Thus, the solution corresponding to the composition M is called maximum boiling Azeotropic
mixture.
• For example, a mixture of nitric acid and water containing68% nitric acid is a maximum boiling
Azeotropic mixture. This mixture boils at 392.5K which is greater than the boiling points of water
(373K) and nitric acid (359K). The distillate obtained on fractional distillation contains same
percentage of nitric acid (68%). Thus, nitric acid and water cannot be separated from the mixture
by fractional distillation.

Ideal Solutions Non ideal solutions with +ve Non ideal solutions with -ve
deviation deviation

1. Obey Raoult’s law at all Do not obey Raoult’s law. Do not obey Raoult’s law
concentration and
temperature.

2. ∆Hmix= 0 ∆Hmix > 0(+ve) ∆Hmix<0. (-ve)

No heat is absorbed or
evolved during dissolution.

3. ∆Vmixing =0 ∆Vmix >0 ∆Vmix<0. (-ve)

Total volume of solution is (+ve)
equal to sum of the
volumes of components.

4. A…..B molecular A…..B molecular interactions A…..B molecular interactions

interactions forces are forces are weaker than A….A forces are stronger than A ….A or
same as A ….A or or B….B molecular interaction B….B molecular interaction forces
B….B molecular interaction forces
forces

5. Escaping tendency of A and Escaping tendency of A and B Escaping tendency of A and B in

B in solution remains the in solution increases. solution decreases
same as in pure
components.
 6. Vapour pressure and Vapour pressure increases and Vapour pressure decreases and
boiling point are same as boiling point decreases. boiling point increases
expected from Raoult’s law

7. Do not form azeotropic Form minimum boiling Form maximum boiling

mixture Azeotropes. Azeotropes.

8. Components can be Components cannot be Components cannot be separated

separated by fractional separated by fractional by fractional distillation
distillation distillation.

COLLIGATIVE PROPERTIES AND MOLECULAR MASS OF NON-VOLATILE SOLUTES.

The properties of solutions which depend only upon the number of solute particles (molecules or ions)
but not on the chemical nature of the solute in the solution are called colligative properties.

There are four colligative properties. They are:

𝑷𝒐𝑨 −𝐏
1. Relative lowering of vapour pressure ( )
𝑷𝒐𝑨
2. Elevation of boiling point (∆Tb)
3. Depression of freezing point(∆Tf)
4. Osmotic pressure. (𝜋)

[I] RELATIVE LOWERING OF VAPOUR PRESSURE.

According to Raoult’s law, the relative lowering of vapour pressure of a solution containing non-volatile
solute is equal to mole fraction of the solute.
𝑷𝒐𝑨 −𝐏 𝑛𝐵
( 𝑷𝒐𝑨
) = XB = 𝑛𝐴 +𝑛𝐵

Relative lowering of vapour pressure depends only on the mole fraction of the solute which in turn
depends on the number of moles of solute dissolved in a definite amount of solvent. Thus, relative
lowering of vapour pressure is a colligative property.

Determination of Molecular Mass

𝑷𝒐𝑨 −𝐏
According to Raoult’s law, ( 𝑷𝒐𝑨
) = XB

𝑷𝒐𝑨 −𝐏 𝑛𝐵
( 𝑷𝒐𝑨
) =𝑛
𝐴 +𝑛𝐵

𝒘𝑩
𝑷𝒐𝑨 −𝐏 𝑴𝑩
( 𝑷𝒐𝑨
) = 𝒘𝑨 𝒘𝑩
+
𝑴𝑨 𝑴𝑩

Knowing the values of vapour pressure solution (P), vapour pressure of pure solvent( 𝑷𝒐𝑨 ), mass of
solute (𝑤𝐵 ), mass of solvent (𝑤𝐴 ), and molecular mass of solvent (𝑀𝐴 ),molecular mass of solute (𝑀𝑩 )
can be calculated.
For a dilute solution, the amount of solute is much less as compared to that of solvent. Hence 𝒏𝑩 in
the denominator can be neglected.

ie 𝑛𝐴 + 𝑛𝐵 ≈ 𝑛𝐴
𝑤𝐵
𝑷𝒐𝑨 −𝐏 𝑀𝐵
( 𝑷𝒐𝑨
) = 𝑤𝐴
𝑀𝐴

𝑷𝒐
𝑨 𝒘𝑩 𝑴𝑨
MB = 𝑷𝒐 −𝐏 x 𝒘𝑨
𝑨

***Measurement of Relative lowering of vapour pressure.

Ostwald and Walker’s dynamic method.

Loss in mass of solution bulbs ∝ P

Loss in mass of solvent bulbs ∝ (𝑷𝒐𝑨 -P)

Gain in mass of CaCl2 tubes or U-tubes ∝ 𝑷𝒐𝑨

𝑷𝒐𝑨 −𝐏 𝐿𝑜𝑠𝑠 𝑖𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑏𝑢𝑙𝑏𝑠
(𝑷𝒐𝑨
) = 𝐺𝑎𝑖𝑛 𝑖𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑎𝐶𝑙2 𝑡𝑢𝑏𝑒𝑠

𝐿𝑜𝑠𝑠 𝑖𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑏𝑢𝑙𝑏𝑠

= 𝐿𝑜𝑠𝑠 𝑖𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑏𝑢𝑙𝑏𝑠 +𝑙𝑜𝑠𝑠 𝑖𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑏𝑢𝑙𝑏𝑠

[II] ELEVATION OF BOILING POINT – EBULLIOSCOPY.

Boiling point.

The temperature at which the vapour pressure of a liquid becomes equal to atmospheric pressure is
called boiling point of the liquid

For example, water boils at 373 K. At this temperature vapour pressure of water becomes equal to 1
atmosphere.

Elevation of Boiling point (∆Tb)

We know that the vapour pressure of a liquid gets lowered when a non-volatile solute is dissolved in it.
Consequently, boiling point of a solution is always higher than that of the pure solvent. The addition of
non-volatile solute to a pure solvent increases its boiling point.

The increase in boiling point of a solvent on the addition of non-volatile solute to it is called elevation of
boiling point.

If T°b is the boiling point of pure solvent and Tb that of solution the difference (Tb - T°b) is called
elevation of boiling point (∆Tb)

∆Tb = (Tb - T°b). ……………. (1)

From the graph, it is clear that elevation of boiling point is directly proportional to lowering of vapour
pressure

∆Tb α (𝑷𝒐𝑨 - P)

But from Raoult’s law, (𝑷𝒐𝑨 - P) α XB

Therefore, ∆Tb α XB

∆Tb = K x XB (K= constant)

𝑛𝐵
∆Tb = K x 𝑛
𝐴 +𝑛𝐵

For dilute solution nB is very small and 𝑛𝐴 + 𝑛𝐵 ≈ 𝑛𝐴 .

𝑛
Therefore ∆Tb = K x 𝑛𝐵
𝐴

But nA = wA/MA
𝑛𝐵 𝑀𝐴
Hence, ∆Tb = K x 𝑤𝐴

𝑛
If mass of solvent wA is expressed in kg, 𝑤𝐵 = m, molality of the solution.
𝐴

Thus, ∆Tb = K x MA x m

Since , K x MA = kb, a new constant, the above equation becomes

∆Tb = kb x m ………… 2

kb is known as molal elevation constant or ebullioscopic constant.

molal elevation constant or ebullioscopic constant

If m=1, kb = ∆Tb
Thus, molal elevation constant is defined as ' the elevation of boiling point when the molality of the
solution is unity.

OR

It is the elevation of boiling point of a one molal solution.

OR

It is the elevation of boiling point of a solution containing one gram mole of solute dissolved in one kg
or 1000g of solvent.

Units of kb.

1. K kg mol -1
2. K molal -1
3. °C molal -1

Molal elevation constants of some solvents are given below.

Water = 0.52 K kg mol -1

Ethanol = 1.2 Kkgmol -1

Benzene = 2.53 K kg mol -1

Important generalization

1. Elevation of boiling point is directly proportional to molality of the solution.

2. Elevation of boiling point is a colligative property, but boiling point is not a colligative
property. This is because, elevation in boiling point depends upon molality of solution which in
turn depends upon the number of moles of solute in the solution.
3. Equimolal solutions of different non-volatile and non-electrolytic solution dissolved in the same
solvent boil at the same temperature and hence exhibit same elevation of boiling point.

Calculation of molecular mass of non-volatile solute on the basis of ∆Tb.

∆Tb = kb x m
𝑛
= kb x( 𝑤𝐵 )x 1000
𝐴

𝑤𝐵
= kb x x 1000
𝑀𝐵 𝑤𝐴

𝟏𝟎𝟎𝟎 𝒌𝒃 𝒘𝑩
∆Tb = 𝑴𝑩 𝒘𝑨
……………………………. 3

OR
𝟏𝟎𝟎𝟎 𝒌𝒃 𝒘𝑩
Molecular mass of solute, MB = ∆𝐓𝒃 𝒘𝑨
……………………… 4
[III] DEPRESSION OF FREEZING POINT (CRYOSCOPY) (∆Tf)

Freezing Point: The temperature at which the liquid and solid forms of a substance can exist together in
equilibrium is called freezing point.

At freezing point vapour pressure of the liquid form is equal to the vapour pressure of its solid form.

When a non-volatile solute is dissolved in a solvent, the freezing point of the solvent gets lowered.

The decrease in freezing point of a solvent caused by the addition of a non-volatile solute to it is

called depression of freezing point. (∆Tf).

∆Tf = T°f - Tf

From the graph, it is clear that depression of freezing point is directly proportional to lowering of vapour
pressure.

∆Tf α (𝑷𝒐𝑨 -P)

But from Raoult’s law, (𝑷𝒐𝑨 -P) α XB

Therefore, ∆Tf α XB

∆Tf = K x XB (K= constant)

𝑛𝐵
∆Tf = K x 𝑛
𝐴 +𝑛𝐵

For dilute solution nB is very small and 𝑛𝐴 + 𝑛𝐵 ≈ 𝑛𝐴 .

𝑛
Therefore ∆Tf = K x 𝑛𝐵
𝐴

But nA = wA/MA
 𝑛𝐵 𝑀𝐴
Hence, ∆Tf = K x 𝑤𝐴

𝑛𝐵
If mass of solvent wA is expressed in kg, = m, molality of the solution.
𝑤𝐴

Thus, ∆Tf = K x MA x m

Since , K x MA = kf, a new constant, the above equation becomes;

∆Tf = kf x m ………… 2

kf is known as molal depression constant or cryoscopic constant.

molal depression constant or cryoscopic constant

If m=1, kf = ∆Tf

Thus, molal depression constant is defined as ' the depression of freezing point when the molality of
the solution is unity.

OR

It is the depression of freezing point of a one molal solution.

OR

It is the depression of freezing point of a solution containing one gram mole of solute dissolved in one
kg or 1000g of solvent.

Units of kf.

1. K kg mol -1
2. K molal -1
3. °C molal -1

Molal depression constants of some solvents are given below.

Water = 1.86 K kg mol-1

Ethanol = 1.99 K kg mol-1

Benzene = 5.12 K kg mol-1

Important generalization

1. Depression of freezing point is directly proportional to molality of the solution.

2. Depression of freezing point is a colligative property, but freezing point is not a colligative
property. This is because, depression of freezing point depends upon molality of solution which
in turn depends upon the number of moles of solute in the solution.
3. Equimolal solutions of different non-volatile and non-electrolytic solution dissolved in the same
solvent freeze at the same temperature and hence exhibit same depression of freezing point.
Calculation of molecular mass of non-volatile solute on the basis of ∆Tf

∆Tf = kf x m
𝑛𝐵
= kf x( )x 1000
𝑤𝐴

𝑤𝐵
= kf x x 1000
𝑀𝐵 𝑤𝐴

𝟏𝟎𝟎𝟎 𝒌𝒇 𝒘𝑩
∆Tf = ……………………………. 3
𝑴𝑩 𝒘𝑨

OR
𝟏𝟎𝟎𝟎 𝒌𝒇 𝒘𝑩
Molecular mass of solute, MB = ……………………… 4
∆𝐓𝒇 𝒘𝑨

Relation between ∆Tb and ∆Tf

∆Tf = kf x m and ∆Tb = kb x m

∆𝐓𝐛 𝒌𝒃
For a solution of same molality, ∆𝐓𝐟
= 𝒌𝒇
………………………… 5

Note:

Anti-freeze solutions.

Water is used in radiators of vehicles. If the vehicle is to be used in places where the temperature is less
than zero, then water would freeze in the radiators. To avoid this problem, certain substances are used
in radiators so that water does not freeze at low temperature in radiators. These are called anti-freeze
solutions. Ethylene glycol in water is commonly used in car radiators which lowers the freezing point of
water. Freezing point can be lowered to the desired extent by changing the amount of ethylene glycol.

Common salt (NaCl) or calcium chloride (CaCl2) are used to clear snow on the roads. This is because they
depress the freezing point of water to such an extent that water cannot freeze to form ice. Hence, it
melts off easily at the prevailing temperature.

[IV] OSMOTIC PRESSURE

Semipermeable membrane.

A membrane which allows the passage of only solvent molecules not of solute particles is called semi
permeable membrane

Examples of natural semi permeable membrane:

1. parchment paper
2. animal and plant cell membranes
3. egg membranes
Examples of artificial semipermeable membranes:

1. Copper ferrocyanide
2. Silicates of Fe, Co and Ni
3. Calcium phosphate

OSMOSIS

The spontaneous flow of solvent molecules through a semipermeable membrane from a pure solvent to
a solution or from a dilute solution to a concentrated solution is called osmosis.

OSMOTIC PRESSURE ( π)

Osmotic pressure of a solution at a particular temperature is defined as the hydrostatic pressure

which builds up on the solution when the solution is separated from the solvent by a semi permeable
membrane and which is just sufficient to stop the phenomenon of osmosis.

OR

The external pressure which should be applied to the solution to stop the phenomenon of osmosis ie
to stop the flow of solvent into solution when the two are separated by a semi permeable membrane
is called osmotic pressure.
REVERSE OSMOSIS

When the pressure applied on the solution is greater than the osmotic pressure of the solution, the
solvent starts flowing from solution into the solvent through the semi permeable membrane. This is
known as reverse osmosis.

Applications:

it is used for the desalination (removal of salts) of seawater or hard water.

Difference between osmosis and diffusion

Osmosis Diffusion

1. Semipermeable membrane is required No semipermeable membrane is required

2. Only solvent molecules move Both solvent and solute molecules move

3. Can be stopped or reversed Cannot be stopped or reversed

4. Shown only by solutions. Shown by solutions and gases

5. Net flow of solvent takes place from Solute particles move from higher
lower concentration to higher concentration to lower concentration and
concentration. solvent particles move from lower to higher
concentration.

Laws of Osmotic Pressure

1. Boyle – van’t Hoff law.

The osmotic pressure of a dilute solution is directly proportional to its concentration at constant
temperature.

π α C at constant temperature [C is concentration in mol/litre]

But C = n/V Therefore,

𝟏
πα𝑽
2. Charles - van't Hoff law.

Osmotic Pressure of a dilute solution is directly proportional to the absolute temperature if

concentration of solution is kept constant.

π α T (at constant Concentration)

3. Avogadro – van’tHoff law.

Equal volumes of solutions contain equal number of moles of solute provided their temperature
and osmotic pressure are same.

Van’tHoff’s equation.

According to Boyle – van’t Hoff law.

παC

According to Charles – van’t Hoff law.

παT

Combining both laws

παCT

π = constant x C T

Value of constant is equal to universal gas constant.

π = R x C T or

π = CRT …………1

𝑛𝐵
But C = 𝑉
𝑛𝐵
Therefore, π= 𝑉
RT

Or

πV = nB RT ………..2

Equations 1 and 2 are called van’t Hoff ‘s equations for solutions.

Calculation of Molecular Mass Using Osmotic Pressure.
wB
πV = nB RT but nB = MB

w
Therefore, πV = M B RT …………………….3
B

𝒘𝑩
Molecular Mass of solute, MB = RT ……………………..4
𝛑𝐕

Note: Osmotic Pressure method is widely used for measuring the molecular mass of proteins, polymers
and other macro molecules because of the following reasons.

i) ∆Tb and ∆Tf values are very small and cannot be measured accurately but osmotic
pressure can be measured accurately at room temperature even for dilute solutions.
ii) Elevation of boiling point method cannot be used because proteins are not stable at
high temperature.

Isotonic Solutions.

The solutions having same osmotic pressure at the same temperature are called isotonic solutions.

When Isotonic solutions are separated by a semipermeable membrane no osmosis takes place.

π1= C1RT for solution1

π2= C2RT. For solution2.

If π1= π2 then C1 = C2
n1 n2
Then =
V1 V2

𝐰𝟏 𝐰𝟐
𝑶𝒓 𝐌𝟏 𝐕𝟏
=𝐌 ………………………. 4
𝟐 𝐕𝟐

If volumes of solutions are equal,

𝐰𝟏 𝐰
= 𝐌𝟐 ………………………..5
𝐌𝟏 𝟐

w1 = mass of solute1

w2 = mass of solute2

M1 = Molecular mass of solute1

M2 = Molecular mass of solute2

Note: A 0.91%(mass/volume) solution of NaCl is Isotonic with fluids in RBC. In this solution, blood
corpuscles neither shrink nor swell. Therefore, medicines are mixed with saline water before being
injected into veins. If the medicine is not isotonic with blood fluids, the blood corpuscles may swell or
shrink due to osmosis.

Hypertonic and Hypotonic Solutions.

When a solution has a higher osmotic pressure as compared to some other solution, it is termed as
hypertonic solution.

A hypertonic solution is more concentrated in comparison to a given solution.

A cell placed in a hypertonic solution will shrink (undergo plasmolysis) due to the movement of water
out of the cell through osmosis

When a solution has a lower osmotic pressure as compared to some other solution, it is termed as
hypotonic solution.

A hypotonic solution is less concentrated in comparison to a given solution.

A cell placed in a hypotonic solution will swell (become turgid) due to the movement of water into the
cell.

High Sodium & Osmotic Pressure:

When you eat a lot of salt (sodium), the sodium concentration in your blood increases, creating a higher
osmotic pressure. This high osmotic pressure in the bloodstream draws water from the cells and tissues
where sodium concentrations are lower, trying to equalize the concentration of solutes. The excess
water that is drawn into the bloodstream is retained, which can lead to swelling or edema.

Conditions for Accurate measurement of Molecular Mass:

1. The solute should be non-volatile.

2. The solution should be dilute.
3. The solute should be a non-electrolyte, ie the solute should not undergo association or
dissociation in solution.

Abnormal Molecular Masses.

When the experimentally observed molecular mass of a solute determined on the basis of colligative
properties is found to be different from the normal molecular mass as expected from its chemical
formula, the observed molecular mass is called Abnormal Molecular mass.

Cause of Abnormal Molecular Mass.

Molecular mass of a solute is inversely proportional to colligative property. But colligative property
is directly proportional to the number of particles of solute in the solution. Thus, molecular mass of
solute is inversely proportional to the number of particles of solute in the solution. Abnormal
Molecular Mass of a solute is due to association or dissociation of solute particles in the solvent. This
is because, due to association or dissociation solute molecules, the number of particles of solute in
the solution changes. This leads to abnormal Molecular Mass.
Association.

Certain solutes undergo association in solution. Two or molecules may combine to form aggregates
(dimer, trimer……). When the solute undergoes association, the number of particles of solute in the
solution decreases. Since the magnitude of colligative properties depends on the number of
particles, the observed value of colligative properties also decreases. But the molecular mass of
solute is inversely proportional to the observed value of colligative property. Hence observed
molecular mass will be higher than the normal or expected molecular mass. Thus association of
solute particles always leads to higher value of observed molecular mass.

For example, the observed molecular mass of benzoic acid in benzene ( in non-polar solvents)
obtained on the basis of colligative property is almost double the normal molecular mass.

Explanation: When benzoic acid is dissolved in benzene, two molecules of it associate together to
form a dimer.

2C6H5-COOH ----→ (C6H5-COOH)2 (dimer)

The formation of dimer reduces the number of solute particles to one half of the normal number of
particles. This leads to decreases in colligative property of the solution to one half of its normal
colligative property. But molecular mass is inversely proportional to the colligative property of the
solution. Therefore, the observed molecular mass will be almost double the normal molecular mass.

For the same reason, the observed molecular mass of acetic acid dissolved in benzene is almost
double the normal molecular mass.

Dissociation.

Certain solutes(electrolytes) undergo dissociation in solution. When the solute undergoes

dissociation, the number of particles of solute in the solution increases. Since the magnitude of
colligative properties depends on the number of particles, the observed value of colligative
properties also increases. But the molecular mass of solute is inversely proportional to the observed
value of colligative property. Hence observed molecular mass will be lower than the normal or
expected molecular mass. Thus, dissociation of solute particles always leads to lower value of
observed molecular mass.

For example, the observed molecular mass of sodium chloride (AB type) obtained on the basis of
colligative property is almost half of the normal molecular mass.

Explanation: Sodium chloride is a strong electrolyte and it dissociates in water as follows.

NaCl → Na+ + Cl-

If the dissociation is 100%, the number of particles becomes double. As a result, the magnitude of
colligative property also becomes double. Since molecular mass of a solute is inversely proportional
to the colligative property, the observed molecular mass of the solute will be half of its normal
molecular mass.

Similarly, the observed molecular mass of K2SO4 (AB2 or A2B type), obtained on the basis of
colligative property is almost one-third of its normal molecular mass.
Explanation: Potassium sulphate is a strong electrolyte and it dissociates in water as follows.

K2SO4 → 2K+ + SO42-

If the dissociation is 100%, the number of particles becomes three times. As a result, the magnitude
of colligative property also becomes three times. Since molecular mass of a solute is inversely
proportional to the colligative property, the observed molecular mass of the solute will be one-third
of its normal molecular mass.

van’tHoff factor

It is a term to express the extend of association or dissociation of a solute in the solution.

It is defined as the ratio of the observed (experimental) value of colligative property to the normal
(calculated) value of the colligative property.

𝐎𝐛𝐬𝐞𝐫𝐯𝐞𝐝 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐜𝐨𝐥𝐥𝐢𝐠𝐚𝐭𝐢𝐯𝐞 𝐩𝐫𝐨𝐩𝐞𝐫𝐭𝐲 ∆𝒐𝒃𝒔.

van’tHoff factor (i) = =
𝐍𝐨𝐫𝐦𝐚𝐥 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐜𝐨𝐥𝐥𝐢𝐠𝐚𝐭𝐢𝐯𝐞 𝐩𝐫𝐨𝐩𝐞𝐫𝐭𝐲 ∆𝒏𝒐𝒓.

𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐚𝐟𝐭𝐞𝐫 𝐚𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐫 𝐝𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧

=
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐢𝐜𝐥𝐞𝐬 𝐛𝐞𝐟𝐨𝐫𝐞 𝐚𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐫 𝐝𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧

𝐎𝐛𝐬𝐫𝐯𝐞𝐝 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

=
𝐍𝐨𝐫𝐦𝐚𝐥 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐢𝐜𝐥𝐞𝐬

𝐍𝐨𝐫𝐦𝐚𝐥 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐦𝐚𝐬𝐬

=
𝐎𝐛𝐬𝐞𝐫𝐯𝐞𝐝 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐦𝐚𝐬𝐬

• When the solute remains unchanged in solution, i = 1

• When the solute undergoes association in solution, i < 1
• When the solute undergoes dissociation in solution i > 1

Modified equations for Colligative properties.

Inclusion of van’tHoff factor (i) modifies the equations for colligative properties and molecular mass.

For colligative properties:

𝑷𝒐𝑨 −𝐏 𝑷𝒐𝑨 −𝐏 𝒏𝑩
1. ( 𝑷𝒐𝑨
) = i XB or ( 𝑷𝒐𝑨
) = 𝒊𝒏
𝑨 +𝒏𝑩

𝟏𝟎𝟎𝟎 𝒌𝒃 𝒘𝑩
2. ∆Tb = i kb x m or ∆Tb = 𝒊 𝑴𝑩 𝒘𝑨

𝟏𝟎𝟎𝟎 𝒌𝒇 𝒘𝑩
3. ∆Tf = i kf x m or ∆Tf = 𝒊 𝑴𝑩 𝒘𝑨

4. π = i CRT or πV = i nB RT
 For molecular mass of non-volatile solute.
𝑷𝒐
𝑨 𝒘𝑩 𝑴𝑨
1. MB = 𝒊 𝑷𝒐 −𝑷 x 𝒘𝑨
𝑨

𝟏𝟎𝟎𝟎 𝒌𝒃 𝒘𝑩
2. MB = 𝒊 ∆𝐓𝒃 𝒘𝑨

𝟏𝟎𝟎𝟎 𝒌𝒇 𝒘𝑩
3. MB = 𝒊 ∆𝐓𝒇 𝒘𝑨

𝐰
4. MB = 𝒊 𝛑𝐕𝐁 RT

Calculation of degree of dissociation (𝜶)

Degree of dissociation is the fraction of the total number of moles of solute which undergoes
dissociation.
Number moles of solute dissociated
Degree of dissociation (𝜶) = Number moles of solute taken

Suppose an electrolytic solute A dissociates in solution to give a total of `n` ions at equilibrium.

A ⇋ n1B + n2C ; where n1+n2 = n.

If we start with 1mole of A,(Normal number of particles)

The amount of solute dissociated from 1mole is = 𝛼 (the degree of dissociation).

Number of moles of A undissociated = 1- 𝛼

Number of moles of ion in the solution = n 𝛼

Total number of particles in the solution = 1- 𝛼 + n 𝛼 (observed number of particles)
𝐎𝐛𝐬𝐞𝐯𝐞𝐝 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
van’tHoff factor (i) = 𝐍𝐨𝐫𝐦𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐢𝐜𝐥𝐞𝐬

1− 𝛼 + n 𝛼
𝒊=
𝟏

𝐓𝐡𝐮𝐬 𝐟𝐨𝐫 𝐝𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧, 𝐯𝐚𝐧′ 𝐭𝐇𝐨𝐟𝐟 𝐟𝐚𝐜𝐭𝐨𝐫 , 𝒊 = 1- 𝜶 + n 𝜶 ------------------------1

Or 𝒊 - 1 = n 𝜶 - 𝜶 = 𝜶 (n-1)
𝒊−𝟏
OR Degree of dissociation , 𝜶 = ------------------- 2
𝒏−𝟏

𝑴𝒏𝒐𝒓.
−𝟏
𝑴𝒏𝒐𝒓. 𝑴𝒐𝒃𝒔.
But 𝒊= 𝑴𝒐𝒃𝒔.
Therefore 𝜶 =
𝒏−𝟏
 𝑴𝒏𝒐𝒓 −𝑴𝒐𝒃𝒔
Or 𝜶 = 𝑴𝒐𝒃𝒔 (𝒏−𝟏)
………………………………. 3

Degree of Association. (𝜶)

Degree of association is the fraction of the total number of moles of solute which undergoes
association.
Number moles of solute associated
Degree of association (𝜶) = Number moles of solute taken

Suppose, n molecules a solute A associates in solution to give an aggregate An at equilibrium.

nA ⇋ An

If we start with 1mole of A,(Normal number of particles):

The amount of solute associated from 1mole is = 𝛼 (the degree of association).

Number of moles of A undissociated = 1- 𝛼

𝛂
Number of moles of particles of An in the solution =
𝐧
𝛂
Total number of particles in the solution = 1- 𝛼 + 𝐧
(observed number of particles)
𝐎𝐛𝐬𝐞𝐯𝐞𝐝 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
van’tHoff factor (i) =
𝐍𝐨𝐫𝐦𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐢𝐜𝐥𝐞𝐬
𝛂
1− 𝛼 +
𝐧
𝒊= 𝟏
𝛂
𝐓𝐡𝐮𝐬 𝐟𝐨𝐫 𝐚𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧, 𝐯𝐚𝐧′ 𝐭𝐇𝐨𝐟𝐟 𝐟𝐚𝐜𝐭𝐨𝐫 , 𝒊 = 1- 𝜶 + ------------------------ 4
𝐧

𝛂 𝟏
Or 𝜶- = 1- 𝒊 Or 𝜶(1 - ) = 1- 𝒊
𝐧 𝐧

𝟏−𝒊
Or degree of association 𝜶= 𝟏 ……………………………..5
𝟏−
𝐧
𝒏(𝟏−𝒊)
Or degree of association 𝜶= ……………………………6
𝒏−𝟏

𝑴𝒏𝒐𝒓.
𝒏(𝟏− )
𝑴𝒏𝒐𝒓. 𝑴𝒐𝒃𝒔.
But 𝒊= 𝑴𝒐𝒃𝒔.
Therefore Therefore, 𝜶=
𝒏−𝟏
𝒏 𝑴𝒐𝒃𝒔. −𝑴𝒏𝒐𝒓.
Or 𝜶= [ ] ……………………7
𝒏−𝟏 𝑴𝒐𝒃𝒔