Address: Shalteng, near HP gas agency opposite Iqbal Masjid Contact: 7006814983

Class: 12th Chemistry Chapter 2: Solutions

INTRODUCTION

In normal life we rarely come across pure substances. Most of these are mixtures containing two or more pure
substances. Their utility or importance in life depends on their composition. The air around us is a mixture of
gases primarily oxygen and nitrogen; the water we drink contains very small amounts of various salts dissolved in
it. Our blood is a mixture of different components. Alloys such as brass, bronze, stainless steel, etc. are also
mixtures. In this Unit, we will consider mostly liquid solutions and their properties.

1. SOLUTIONS

1.1 Definition

A solution is a homogeneous mixture of two or more than two components.For example, common salt in water.

1.2 Classification

Solutions which contain two components in it are calledBinary Solutions.

Substances which are used to prepare a solution are calledas Components.
The component that is present in the largest quantity is known as Solvent. Solvent determines the physical state
in which solution exists.
The other component present in lesser quantity in the solution is termed as Solute.
Each component may be solid, liquid or in gaseous state.

1.3 Strength of Solutions

The amount of solute dissolved per unit solution or solvent is called Strength of solution. There are various
methods of measuring strength of a solution.
1. Mass Percentage (%w/w):It represents mass of a component present in 100 g of solution.

𝑀𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑜𝑙

𝑀𝑎𝑠𝑠 % 𝑜𝑓 𝑎 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 = × 100
𝑇𝑜𝑡𝑎𝑙 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙.

2. Volume percentage (%v/v):It represents volume of a component in 100 mL of solution.

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑜𝑙

𝑉𝑜𝑙𝑢𝑚𝑒 % 𝑜𝑓 𝑎 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 = × 100
𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.

3. Mass by volume percentage (%w/v):It represents mass of solute in grams present in 100 mL of solution.

𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔
𝑀𝑎𝑠𝑠 𝑏𝑦 𝑉𝑜𝑙𝑢𝑚𝑒 % = × 100
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑙

4. Parts per Million (ppm):

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠

𝑝𝑝𝑚 = × 106
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑜𝑙.

Concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume.

5. Mole Fraction (x):It represents the moles of a solute present in one mole of solution.

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠

𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑜𝑙.

For example, in a binary mixture, if the number of moles of A and B are 𝑛𝑎 and 𝑛𝑏 respectively, the mole
fraction of A will be
𝑛𝑎
𝜒𝑎 =
𝑛𝑎 + 𝑛𝑏

6. Molarity, M: It represents moles of solute present in 1 L of solution.

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑎𝑙𝑢𝑡𝑒
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦, 𝑀 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙. 𝑖𝑛 𝐿

Units of Molarity are mol/L also represented by ‘M’ or ‘Molar’.

7. Molality, 𝒎:It represents moles of solute present per kg of solvent.

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑎𝑙𝑢𝑡𝑒
𝑀𝑜𝑙𝑎𝑙𝑖𝑡𝑦, 𝑚 =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝐾𝑔

Units of molality are mol/kg which is also represented by ‘m’ or ‘molal’.

8. Normality, N: It represents no. of equivalents of solute present in 1 L of solution.

𝑁𝑜. 𝑜𝑓 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦, 𝑁 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙. 𝑖𝑛 𝐿

𝑊𝑒𝑖𝑔ℎ𝑡
𝑁𝑜. 𝑜𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠, 𝑒𝑞 = 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 (𝑊/𝐸)

𝑀
𝐸= (𝑧 is the valency factor)
𝑧

SOME IMPORTANT RELATIONSHIPS

Dilution Law: If a solution is diluted by adding solvent to it, then the amount of solute remains constant and we
can write: M1V1 = M2V2 and N1V1 = N2V2
Molarity and NormalityNormality = z × Molarity

2. VAPOUR PRESSURE
2.1 Definition: Vapour pressure of a liquid/solution is the pressure exerted by the vapours in equilibrium with
the liquid/solution at a particular temperature.

𝑉𝑎𝑝𝑜𝑢𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∝ 𝑒𝑠𝑐𝑎𝑝𝑖𝑛𝑔 𝑡𝑒𝑛𝑑𝑒𝑛𝑐𝑦

2.2 Vapour pressure of liquid solutions and Raoult’sLaw :

1). Raoult’s law for volatile solutes: Raoult’s law states that for a solution of volatile liquids, the partial vapour
pressure of each component in the solution is directly proportional to its mole fraction.Consider a solution
containing two volatile components 1and 2 with mole fractions 𝜒1 and 𝜒2 respectively. Supposeat a particular
temperature, their partial vapour pressures are𝑝1 and 𝑝2 and the vapour pressure in pure state are𝑝10 and
𝑝20 .Thus, according to Raoult’s Law, for component 1
𝑝1 ∝ 𝜒1
And 𝑝1 = 𝑝10 𝜒1
Similarly, for component 2
𝑝2 = 𝑝20 𝜒2
According to Dalton’s law of partial pressure, the total pressure (𝑝𝑡𝑜𝑡𝑎𝑙 ) over the solution phase in the container
will be the sum of the partial pressures of the components of the solution and is given as :
𝑝𝑡𝑜𝑡𝑎𝑙 = 𝑝1 + 𝑝2
Substituting the values of𝑝1 and𝑝2 , we get
𝑝𝑡𝑜𝑡𝑎𝑙 = 𝑝10 𝜒1 + 𝑝20 𝜒2

𝑝𝑡𝑜𝑡𝑎𝑙 = 𝑝10 (1 − 𝜒2 ) + 𝑝20 𝜒2

𝑝𝑡𝑜𝑡𝑎𝑙 = 𝑝10 + (𝑝20 − 𝑝10 ) 𝜒2

The plot of vapour pressure and mole fraction of an ideal solution at constant temperature. The dashed line I and
II represent the partial pressure of the components. It can be seen from the plot that 𝑝1 and 𝑝2 are directly
proportional to 𝜒1 and 𝜒2 , respectively. The total vapour pressure is given by line marked III in the figure.
Mole fraction in vapour phase: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 and 𝑝2 = 𝑦2 𝑝𝑡𝑜𝑡𝑎𝑙

In general, 𝑝𝑖 = 𝑦𝑖 𝑝𝑡𝑜𝑡𝑎𝑙

2.3 Vapour pressures of solutions of solids in liquids and Raoult’s Law

2). Raoult’s law for non volatile solutes:If a non-volatile solute is added to a solvent to give a solution, the
number of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapour pressure is
also reduced.

The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the
solution, irrespective of its nature.

Raoult’s law in its general form 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.

In a binary solution, let us denote the solvent by 1 and solute by 2. When the solute is non-volatile, only the
solvent molecules are present in vapour phase and contribute to vapour pressure. Let 𝑝1 be the vapour pressure
of the solvent, 𝜒1 be its mole fraction, 𝑝10 be its vapour pressure in the pure state. Then according to Raoult’s law.

𝑝1 ∝ 𝜒1
And 𝑝1 = 𝑝10 𝜒1 = 𝑝𝑡𝑜𝑡𝑎𝑙

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.

2.4 Ideal and Non-ideal solutions:

Ideal solutions :An ideal solution is the solution in which each component obeys Raoult’s law under all
conditions of temperatures and concentrations.

Properties of Ideal solutions :

  ∆𝐻𝑚𝑖𝑥 = 0
 ∆𝑉𝑚𝑖𝑥 = 0
 Intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B.
E.g. solution of benzene and toluene,solution of n-hexane and n-heptane

Non – ideal solutions :When a solution does not obey Raoult’s law over the entire range of concentration, then
it is called non-ideal solution.

Solutions showing positive deviation from Raoult’sLaw :

 Solvent-Solute(A-B) type of force is weaker than Solute-Solute(B-B) & Solvent-Solvent(A-A) forces.

 The vapour pressure is higher than predicted by the law.
 ∆𝐻𝑚𝑖𝑥 > 0
 ∆𝑉𝑚𝑖𝑥 > 0
E.g. ethanol and acetone, carbon disulphide and acetone.
Pressure composition curve for solution showing positive deviation.

Solutions showing negative deviations from Raoult’slaw :

 Solvent-Solute(A-B) type of force is stronger than the other two.

 The vapour pressure is lower than predicted by the law.
 ∆𝐻𝑚𝑖𝑥 > 0
 ∆𝑉𝑚𝑖𝑥 > 0
For example,phenol and aniline, chloroform and acetone etc
Pressure composition curves for solution showing negative deviation.
2.5 Azeotropes:Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil
at a constant temperature.

Minimum boiling azeotrope:The solutions which show a large positive deviation from Raoult’s law form
minimum boiling azeotrope at a specific composition.

For example, ethanol-water mixture containing approximately 95% of ethanol forms an azeotrope with boiling
point 351.15 K.

Maximum boiling azeotrope :The solutions that show large negative deviation from Raoult’s law form maximum
boiling azeotrope at a specific composition. Nitric acid and water mixture containing 68% nitric acid forms an
azeotrope with a boiling point of 393.5 K.

3. SOLUBILITY

3.1 Solubility of a solid in liquid:Solubility of a substance is its maximum amount that can be dissolved in a
specified amount of solvent.

Factors affecting the solubility of a solid in liquid :

1. Nature of solute and solvent :Like dissolves like. For example, While sodium chloride and sugar dissolve readily
in water, naphthalene and anthracene do not. On the other hand, naphthalene and anthracene dissolve readily
in benzene but sodium chloride and sugar do not.

2. Temperature :In a nearly saturated solution If (∆𝐻𝑠𝑜𝑙 > 0), the solubility increases with rise in temperature
and if ((∆𝐻𝑠𝑜𝑙 < 0), the solubility decreases with rise in temperature.

3. Effect of pressure :Does not have any significant effect as solids and liquids are highly incompressible.

3.2 Henry’s law: Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly
proportional to the pressure of the gas.

The most commonly used form of Henry’s law states that “the partial pressure of the gas in vapour phase (𝑝) is
proportional to the mole fraction of the gas (𝑥) in the solution”. This is expressed as:

𝑝 = 𝐾𝐻 𝑥

Here 𝐾𝐻 is the Henry’s law constant.

Characteristics of 𝑲𝑯 :

 𝐾𝐻 is a function of the nature of the gas.

 Higher the value of 𝐾𝐻 at a given pressure, the lower is the solubility of the gas in the liquid.
 𝐾𝐻 values increase with increase of temperature indicating that the solubility of gases increases with
decrease of temperature.

Applications of Henry’s law

1. In the production of carbonated beverages.
2. In the deep-sea diving.
3. For climbers or people at high altitudes.

Raoult’s Law as a special case of Henry’s Law:According to Raoult’s law,

 𝑝𝑖 = 𝑝𝑖0 𝜒𝑖

In the solution of a gas in a liquid, one of the components is so volatile that it exists as a gas. Its solubility
according to Henry’s law,

𝑝 = 𝐾𝐻 𝑥
Thus, Raoult’s law becomes a special case of Henry’s law in which𝐾𝐻 becomes equal to 𝑝𝑖0 .

4. COLLIGATIVE PROPERTIES

The properties that depend on the number of solute particles irrespective of their nature relative to the total
number of particles present in the solution are called colligative properties.

There are four colligative properties:

1. Relative Lowering of vapourPressure
2. Elevation in Boiling Point
3. Depression in freezing point
4. Osmotic pressure

4.1 Relative Lowering of vapour Pressure:When a non-volatile solute is added to a solvent, the vapour pressure
decreases.The lowering of vapour pressure w.r.t. the vapour pressure of the pure solvent is called “Relative
lowering in vapour pressure”.

According to Raoult’sLaw :𝑝1 = 𝜒1 𝑝10

The reduction in the vapour pressure of solvent (∆𝑝1 )given as:

∆𝑝1 = 𝑝10 − 𝑝1 =𝑝10 − 𝜒1 𝑝10 = (1 − 𝜒1 )𝑝10

Knowing that 𝜒2 = 1 − 𝜒1 , equation reduces to∆𝑝1 = 𝜒2 𝑝10

∆𝑝1 𝑝10 −𝑝1

Equation can be wirtten as = = 𝜒2
𝑝10 𝑝10

The expression on the left hand side of the equation as mentioned earlier is called relative lowring of vapour
pressure and is equal to the mole fraction of the solute. The above equation can be written as :

𝑝10 − 𝑝1 𝑛2 𝑛2
0 = (𝑠𝑖𝑛𝑐𝑒 𝜒2 = )
𝑝1 𝑛1 + 𝑛2 𝑛1 + 𝑛2

Here𝑛1 and 𝑛2 are the number of moles of solvent and solute respectively present in the solution. For dilute
solutions 𝑛2 <<𝑛1 , hence neglecting 𝑛2 in the denominator we have

𝑝10 − 𝑝1 𝑛2
=
𝑝10 𝑛1

𝑝10 −𝑝1 𝑤2 𝑀1
Or =
𝑝10
𝑀2 𝑤1
Here 𝑤1 and 𝑤2 are the masses and 𝑀1 and 𝑀2 are the molar masses of the solvent and solute respectively.
4.2 Elevation in Boiling Point:Boiling point of a liquid is the temperature at which the vapour pressure of the
liquid becomes equal to the atmospheric pressure.On addition of non-volatile solute the vapour pressure of the
solvent decreases and therefore, to boil the solution the required temperature will be higher. So, there will be a
rise in the boiling point of the solution.The increase in boiling point ∆𝑇𝑏 = 𝑇𝑏 − 𝑇𝑏0 where𝑇𝑏0 is the boiling point
of pure solvent and 𝑇𝑏 is the boiling point of solution is known as elevation of boiling point.
Expression :∆𝑇𝑏 = 𝐾𝑏 𝑚
𝐾𝑏 is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).
Calculation of molar mass of solute :
𝑤2 /𝑀2 1000 × 𝑤2
𝑚= =
𝑤1 /1000 𝑀2 𝑤1

1000×𝑤
Substituting the value of molality in equation we get∆𝑇𝑏 = 𝐾𝑏 𝑀 𝑤 2
2 1

1000×𝑤
𝑀2 = 𝐾𝑏 ∆𝑇 𝑤 2 𝐾𝑏 : It is defined as the elevation in boiling point when the molality of the solution is
𝑏 1
unity.The unit of 𝐾𝑏 is K kg mol–1
2
𝑅×𝑀1×𝑇𝑏
Determination of 𝐾𝑏 : 𝐾𝑏 = 1000×∆
𝑣𝑎𝑝𝐻

where: R = gas constant (8.314 JK/mol),

𝑇𝑏 = boiling temperature in K,
M1 = Molar mass of solvent in Kg/mol,
∆𝑣𝑎𝑝 𝐻= enthalpy of vapourisation of solvent in J/mol.

The vapour pressure curve for solution lies below the curve for pure water. The diagram shows that Tb denotes
the elevation of boiling point of a solvent in solution.

4.3 Depression in freezing point:The freezing point of a substance may be defined as the temperature at which
the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.When a
non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that
of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.

∆𝑇𝑓 = 𝑇𝑓 − 𝑇𝑓0 where 𝑇𝑓0 is the freezing point of pure solvent and 𝑇𝑓 is its freezing point when non-volatile solute
is dissolvedis known as depression in freezing point.

Expression :∆𝑇𝑓 = 𝐾𝑓 𝑚𝐾𝑓 is known as Freezing Point Depression Constant or Molal Depression Constant or
Cryoscopic Constant.

Calculation of molar mass of solute :

 𝑤2 /𝑀2 1000 × 𝑤2
𝑚= =
𝑤1 /1000 𝑀2 𝑤1

1000×𝑤2
Substituting the value of molality in equation we get∆𝑇𝑓 = 𝐾𝑓 𝑀2𝑤1
𝑀2 = 𝐾𝑓 1000×𝑤
∆𝑇 𝑤
2 𝐾 : It is defined as the elevation in boiling point when the molality of the solution is
𝑓
𝑓 1
unity.The unit of 𝐾𝑓 is K kg mol–1
𝑅×𝑀1×𝑇2𝑓
Determination of 𝐾𝑓 : 𝐾𝑓 = 1000×∆ 𝐻
𝑓𝑢𝑠

where: R = gas constant (8.314 JK/mol),

𝑇𝑓 = boiling temperature in K,

M1 = Molar mass of solvent in Kg/mol,

∆𝑓𝑢𝑠 𝐻 = enthalpy of fusion of solvent in J/mol.

Diagram showing∆𝑇𝑓 depression of the freezing point of a solvent in a solution.

4.4 Osmosis:When a pure solvent and solution are kept with a semipermeable membrane between them then
the solvent particles pass through the membrane from the solvent side to the solution side. This phenomenon is
called “Osmosis”.

The semi-permeable membrane is a membrane that allows only small molecules to pass through and blocks the
larger solute molecules.

Osmotic pressure :The osmotic pressure of a solution is the excess pressure that must be applied to a solution to
prevent osmosis, i.e., to stop the passage of solvent molecules through a semi permeable membrane into the
solution.
The excess pressure equal to osmotic pressure must be applied on the solution to prevent osmosis.

Expression : For dilute solutions, osmotic pressure is proportional to the molarity, 𝐶 of the solution at a given
temperature 𝑇. Thus:

𝜋= 𝐶𝑅𝑇

Here 𝜋 is the osmotic pressure and 𝑅 is the gas constant.

Calculation of molar mass :

𝜋 = 2 (𝑛2 /𝑉) 𝑅𝑇
Here 𝑉 is volume of a solution in litres containing 𝑛2 moles of solute. If 𝑤2 grams of solute, of molar mass, 𝑀2 is
present in the solution, then 𝑛2 = 𝑤2 / 𝑀2 and we can write,
𝑤2 𝑅𝑇 𝑤2 𝑅𝑇
𝜋𝑉 = ⇒ 𝑀2 =
𝑀2 𝜋𝑉

Isotonic solutions :Two solutions having same osmotic pressure at a given temperature are called isotonic
solutions.

Hypotonic solutions :The solution with lower concentration or lower osmotic pressure is known as “Hypotonic”
with respect to more concentrated solution.

Hypertonic solution: The solution with higher concentration or higher osmotic pressure is known as“Hypertonic”
with respect to dilute solution.

Reverse osmosis :If a pressure larger than the osmotic pressure is applied to the solution side, the solvent will
flow from the solution into the pure solvent through the semi permeable membrane. This phenomenon is called
reverse osmosis.

Application :
Desalination of sea water : When pressure more than osmotic pressure is applied, pure water is squeezed out of
the sea water through the membrane.

4.5 Abnormal Molar Masses

When the molecular mass of a substance determined by studying any of the colligative properties comes out to
be different than the theoretically expected value, the substance is said to show abnormal molar mass.
Abnormal Molar Masses are observed:
1. When the solute undergoes association in the solution.
2. When the solute undergoes dissociation in the solution.

van’t Hoff Factor : To calculate extent of association or dissociation, van’t Hoff introduced a factor i, known as
the van’t Hoff Factor.
𝑁𝑜𝑟𝑚𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑖=
𝐴𝑏𝑛𝑜𝑟𝑚𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑐𝑜𝑙𝑙𝑖𝑔𝑎𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦

𝑖=
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑐𝑜𝑙𝑙𝑖𝑔𝑎𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦

𝑇𝑜𝑡𝑎𝑙 𝑛𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑓𝑡𝑒𝑟 𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 (𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛)

=
𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑏𝑒𝑓𝑜𝑟𝑒 𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 (𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛)

Association : Number of particles will always decrease due to association therefore 𝑖 < 1.
𝑛𝐴 → 𝐴𝑛
Let initial particles (𝑛𝑖 ) = 1
Final number( 𝑛𝑓 ) = 1 – 𝛼 + 𝛼/𝑛
van’t Hoff factor, 𝑖 = 𝑛𝑓 /𝑛𝑖 = 1 – 𝛼 + 𝛼/𝑛

Dissociation : The number of particles will always increase due to dissociation and hence 𝑖 > 1.
𝐴𝑛 → 𝑛𝐴
Initial particle = 1
Final particles = 1 – 𝛼 + 𝛼/𝑛
van’t Hoff factor, 𝑖 = 1 – 𝛼 + 𝛼/𝑛

Modified Expressions :

𝑝10 −𝑝1 𝑛2
Relative lowering of vapour pressure of solvent, =𝑖
𝑝10 𝑛1

Elevation of Boiling point, ∆𝑇𝑏 = 𝑖𝐾𝑏 𝑚

Depression of Freezing point, ∆𝑇𝑓 = 𝑖𝐾𝑓 𝑚𝐾𝑓

Osmotic pressure of solution, 𝜋=𝑖𝐶𝑅𝑇