Solutions Chemistry Class 12: Complete Concepts, Formulas, and Exam Notes by PW

The Solutions chapter combines chemical concepts with numerical problems involving concentration, solubility, vapour pressure and colligative properties. To solve these questions, you need to understand the relationship between solute and solvent, use concentration formulas correctly, and identify how temperature, pressure and the number of dissolved particles affect a solution.

PW’s video lessons cover important concepts and calculations from Solutions Chemistry Class 12 for CUET UG 2027 and Board examinations. Use the lecture alongside these notes to revise definitions, understand the formulas and practise numerical questions based on the chapter.

Types of Solutions

A solution is a homogeneous mixture of a solute and a solvent. Solutions are classified according to the physical states of their components. The solvent is the component that acts as the dissolving medium.

Gaseous Solutions

In a gaseous solution, the solvent is a gas. Examples include:

• Gas in gas: Oxygen and nitrogen in the atmosphere.

• Liquid in gas: Chloroform in nitrogen gas.

• Solid in gas: Camphor in nitrogen gas.

Liquid Solutions

In a liquid solution, the solvent is a liquid. Examples include:

• Gas in liquid: Oxygen dissolved in water.

• Liquid in liquid: Ethanol in water.

• Solid in liquid: Sugar or glucose in water.

Dissolved oxygen in water supports respiration in aquatic organisms.

Solid Solutions

In a solid solution, the solvent is a solid. Examples include:

• Gas in solid: Hydrogen in palladium.

• Liquid in solid: Amalgam of mercury with sodium.

• Solid in solid: Copper in gold.

A mixture of mercury with another substance is called an amalgam.

Concentration of a Solution

Concentration indicates the amount of solute present in a given amount of solution. Different concentration terms are used depending on whether the quantities are expressed by mass, volume or number of moles.

1. Mass Percentage

Mass percentage expresses the mass of solute per 100 parts by mass of solution.

Formula:

Mass percentage = (Mass of solute / Mass of solution) × 100

Mass of solution = Mass of solute + Mass of solvent.

For example, a 10% glucose solution by mass contains 10 g of glucose in 100 g of solution. The remaining 90 g is the solvent.

Memory tip: For mass percentage questions, assume 100 g of solution when convenient. The percentage directly gives the mass of solute.

2. Volume Percentage

Volume percentage expresses the volume of solute per 100 parts by volume of solution.

Formula:

Volume percentage = (Volume of solute / Volume of solution) × 100

For example, a 20% ethanol solution by volume contains 20 mL of ethanol per 100 mL of solution.

3. Mass by Volume Percentage

Mass by volume percentage expresses the mass of solute present in a specified volume of solution.

Formula:

Mass by volume percentage = (Mass of solute / Volume of solution) × 100

When using this formula, keep the mass and volume units consistent with the percentage convention specified in the question.

4. Parts Per Million (PPM)

Parts per million is used to express very low concentrations of a solute.

Formula:

PPM = (Parts of solute / Total parts of solution) × 10⁶

For dilute aqueous solutions, concentration may sometimes be expressed in mg/L as an approximation to ppm, depending on the solution's density.

5. Mole Fraction

Mole fraction is the ratio of the number of moles of one component to the total number of moles of all components in the solution.

Formula:

Mole fraction of component i = nᵢ / Total moles

For a binary solution containing components A and B:

Xₐ = nₐ / (nₐ + nᵦ)

Xᵦ = nᵦ / (nₐ + nᵦ)

The sum of mole fractions is always 1:

Xₐ + Xᵦ = 1

Unlike percentage, mole fraction is not multiplied by 100.

Molarity and Molality

Molarity and molality both measure concentration, but they use different denominators. Identifying which quantity is given in a question is important for selecting the correct formula.

Molarity

Molarity is the number of moles of solute present per litre of solution. It is represented by a capital M.

Formula:

Molarity (M) = Moles of solute / Volume of solution in litres

Example: Calculate the molarity of a solution containing 5 g of NaOH in 450 mL of solution.

Molar mass of NaOH = 40 g/mol.

Moles of NaOH = 5 / 40 = 0.125 mol

Volume of solution = 450 mL = 0.450 L

Molarity = 0.125 / 0.450 = 0.278 M

Molality

Molality is the number of moles of solute present per kilogram of solvent. It is represented by a lowercase m.

Formula:

Molality (m) = Moles of solute / Mass of solvent in kilograms

Always convert the mass of the solvent into kilograms before calculating molality.

Difference Between Molarity and Molality

Molarity changes with temperature because the volume of a solution can change. Molality remains constant with temperature because the mass of the solvent does not change.

Mass percentage, ppm and mole fraction are also independent of temperature when the composition of the solution remains unchanged.

Memory tip: Molarity uses the volume of the solution; molality uses the mass of the solvent.

Solubility of Solids in Liquids

Solubility is the maximum amount of solute that dissolves in a specified amount of solvent at a given temperature to form a saturated solution. It depends on the nature of the solute and solvent and on temperature. Pressure generally has little effect on the solubility of solids in liquids.

Like Dissolves Like

The principle of “like dissolves like” means that substances with similar types of intermolecular interactions tend to dissolve in one another.

• Polar substances such as sodium chloride and sugar dissolve in polar water to different extents.

• Non-polar substances such as naphthalene and anthracene do not dissolve readily in water.

• Naphthalene and anthracene dissolve more readily in non-polar solvents such as benzene.

Saturated and Unsaturated Solutions

• Unsaturated solution: A solution in which more solute can dissolve at the same temperature.

• Saturated solution: A solution that contains the maximum amount of dissolved solute possible at a given temperature.

In a saturated solution at equilibrium, dissolution and crystallisation occur at equal rates. This is called dynamic equilibrium.

Effect of Temperature on Solubility

The effect of temperature depends on whether dissolution is endothermic or exothermic.

• Endothermic dissolution: ΔH > 0. Increasing temperature generally increases solubility.

• Exothermic dissolution: ΔH < 0. Increasing temperature generally decreases solubility.

Pressure has no significant effect on the solubility of most solids in liquids because solids and liquids are nearly incompressible.

Solubility of Gases in Liquids

The solubility of a gas in a liquid generally increases with an increase in pressure and decreases with an increase in temperature.

At a constant temperature, gas solubility is proportional to the gas's partial pressure above the solution.

General relationship:

Gas solubility ∝ Pressure

Oxygen is only slightly soluble in water, but dissolved oxygen is essential for aquatic life. Hydrogen chloride (HCl) is highly soluble in water because of favourable interactions between the gas and the polar solvent.

Also Read: CUET Chemistry Important Topics 2026, Unit-Wise Weightage, Exam Pattern

Henry’s Law

Henry’s law states that at a constant temperature, the partial pressure of a gas above a solution is proportional to its mole fraction in the solution.

Formula:

P = KₕX

Where:

• P = Partial pressure of the gas

• X = Mole fraction of the dissolved gas

• Kₕ = Henry’s law constant

At a given temperature, a higher value of Kₕ means that a higher partial pressure is needed to achieve the same mole fraction of dissolved gas. Therefore, gas solubility is inversely related to Kₕ under the conditions described by the law.

For many gases, Kₕ increases with temperature, so gas solubility decreases as temperature rises.

Applications of Henry’s Law

• Carbonated drinks: Carbon dioxide is dissolved under pressure. Opening the container lowers the pressure, allowing dissolved CO₂ to escape.

• Scuba diving: During rapid ascent, dissolved nitrogen can form bubbles in the blood and tissues, potentially causing decompression sickness, commonly called the bends.

• High altitudes: Lower atmospheric pressure reduces the partial pressure of oxygen. This can reduce oxygen availability to the body and contribute to hypoxia.

Raoult’s Law and Vapour Pressure

Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid at a given temperature. When volatile liquids are mixed, the total vapour pressure depends on the vapour pressures and mole fractions of the individual components.

For a binary solution containing two volatile liquids, Raoult’s law gives:

P₁ = P₁⁰X₁

P₂ = P₂⁰X₂

Total vapour pressure:

P(total) = P₁ + P₂

Therefore:

P(total) = P₁⁰X₁ + P₂⁰X₂

Where:

• P₁ and P₂ = Partial vapour pressures of components 1 and 2

• P₁⁰ and P₂⁰ = Vapour pressures of the pure components

• X₁ and X₂ = Mole fractions of the components in the liquid solution

A component with a higher pure vapour pressure is more volatile. For a non-ideal mixture, however, the actual vapour composition also depends on deviations from Raoult’s law.

Vapour Pressure of a Solution Containing a Non-Volatile Solute

When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases.

For solvent 1:

P₁ = P₁⁰X₁

The solvent's mole fraction is less than 1 when solute is present, so its vapour pressure is lower than that of the pure solvent at the same temperature.

Ideal and Non-Ideal Solutions

Ideal Solutions

An ideal solution obeys Raoult’s law throughout the entire range of composition.

For an ideal solution:

ΔH(mix) = 0

ΔV(mix) = 0

This means that there is no enthalpy change or volume change on mixing. The interactions between unlike molecules are comparable to those between like molecules.

Examples include n-hexane mixed with n-heptane and benzene mixed with toluene.

Non-Ideal Solutions

Non-ideal solutions deviate from Raoult’s law because the interactions between unlike molecules differ from those between like molecules.

Azeotropes

An azeotrope is a liquid mixture whose vapour has the same composition as the liquid at its boiling point. Such mixtures cannot be separated into their components by ordinary fractional distillation.

Colligative Properties

Colligative properties depend on the number of solute particles present in a solution rather than their chemical identity, provided the solution behaves according to the assumptions of the relevant equations.

The four main colligative properties are:

1. Relative lowering of vapour pressure

2. Elevation of boiling point

3. Depression of freezing point

4. Osmotic pressure

1. Relative Lowering of Vapour Pressure

For a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute under the conditions of Raoult’s law.

Formula:

Relative lowering of vapour pressure = ΔP / P⁰ = X₂

Where:

• ΔP = P⁰ − P

• P⁰ = Vapour pressure of the pure solvent

• P = Vapour pressure of the solution

• X₂ = Mole fraction of the solute

For a dilute solution:

ΔP / P⁰ ≈ (w₂ × M₁) / (w₁ × M₂)

Where:

• w₁ = Mass of solvent

• w₂ = Mass of solute

• M₁ = Molar mass of solvent

• M₂ = Molar mass of solute

2. Elevation of Boiling Point

Adding a non-volatile solute lowers the vapour pressure of a solvent. Therefore, the solution must be heated to a higher temperature before its vapour pressure equals the external pressure and boiling occurs.

Formula:

ΔTᵦ = Tᵦ − Tᵦ⁰

For a dilute solution:

ΔTᵦ = Kᵦm

Where:

• ΔTᵦ = Elevation in boiling point

• Tᵦ = Boiling point of the solution

• Tᵦ⁰ = Boiling point of the pure solvent

• Kᵦ = Molal elevation constant

• m = Molality

The molar mass of the solute can be calculated using:

M₂ = (Kᵦ × 1000 × w₂) / (ΔTᵦ × w₁)

Here, w₂ is the mass of solute in grams, w₁ is the mass of solvent in grams, and M₂ is the molar mass of the solute, provided the units are used consistently.

3. Depression of Freezing Point

A solution containing a dissolved solute generally freezes at a lower temperature than the pure solvent.

Formula:

ΔT𝒇 = T𝒇⁰ − T𝒇

For a dilute solution:

ΔT𝒇 = K𝒇m

Where:

• ΔT𝒇 = Depression in freezing point

• T𝒇⁰ = Freezing point of the pure solvent

• T𝒇 = Freezing point of the solution

• K𝒇 = Molal depression constant

• m = Molality

The molar mass of the solute can be calculated using:

M₂ = (K𝒇 × 1000 × w₂) / (ΔT𝒇 × w₁)

Also Check: PW CUET Wallah YouTube Channel

4. Osmotic Pressure

Osmosis is the movement of solvent through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.

Osmotic pressure is the minimum external pressure that must be applied to a solution to stop osmosis.

For a dilute solution:

π = CRT

Alternatively:

π = nRT / V

Where:

• π = Osmotic pressure

• C = Molar concentration

• n = Moles of solute

• R = Gas constant

• T = Absolute temperature in kelvin

• V = Volume of solution

The molar mass of the solute can be calculated using:

M₂ = (w₂ × R × T) / (π × V)

Use compatible units for R, T, π and V.

Biological and Practical Applications of Osmosis

• Isotonic solutions: Have the same effective osmotic pressure as the reference solution under the stated conditions.

• Hypertonic solutions: Can cause cells to lose water and shrink.

• Hypotonic solutions: Can cause cells to gain water and swell.

• Reverse osmosis: Applies pressure greater than the osmotic pressure to force solvent through a semipermeable membrane, a process used in desalination.

A 0.9% sodium chloride solution is commonly described as isotonic with blood under standard physiological conditions.

Abnormal Molar Mass and the van’t Hoff Factor

Some solutes associate or dissociate when dissolved. This changes the number of particles in solution and can cause the molar mass calculated from colligative properties to differ from the expected value.

• Dissociation: One solute unit separates into two or more particles. The number of particles increases, and the apparent molar mass is generally lower.

• Association: Two or more solute units combine to form larger particles. The number of particles decreases, and the apparent molar mass is generally higher.

van’t Hoff Factor

The van’t Hoff factor, represented by i, accounts for the change in the number of solute particles due to association or dissociation.

Formula:

i = Normal molar mass / Abnormal molar mass

It can also be understood as the ratio of the actual number of particles in solution to the number expected if no association or dissociation occurred, for the same amount of solute.

For dissociation, i > 1.

For association, i < 1.

Modified Colligative Property Equations

When association or dissociation occurs, the colligative property equations are modified using the van’t Hoff factor.

Relative lowering of vapour pressure:

ΔP / P⁰ = iX₂

Elevation of boiling point:

ΔTᵦ = iKᵦm

Depression of freezing point:

ΔT𝒇 = iK𝒇m

Osmotic pressure:

π = i(nRT / V)

Examples of Dissociation

Potassium chloride dissociates into potassium and chloride ions:

KCl → K⁺ + Cl⁻

For complete dissociation, one formula unit produces two ions, so i approaches 2.

Potassium sulphate dissociates into potassium and sulphate ions:

K₂SO₄ → 2K⁺ + SO₄²⁻

For complete dissociation, one formula unit produces three ions, so i approaches 3.

Actual values of i may differ from these ideal values depending on concentration and the extent of dissociation.

Watch the Solutions Chemistry Class 12 One-Shot Video to revise important concepts, formulas and numerical problems from the chapter for CUET UG 2027 and board examinations.

The Solutions chapter brings together concentration calculations, solubility principles, vapour pressure and the effects of dissolved particles on boiling point, freezing point and osmotic pressure. For revision, focus on selecting the correct formula, converting units carefully and understanding the conditions under which each equation applies. Use PW’s one-shot lecture with these notes, then practise numerical questions and CUET PYQs to prepare for CUET UG 2027 and Class 12 Board examinations.