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Unit-I: Solutions JKBOSE Class 12 Chemistry — Complete Study Material

🎯 Board Exam Focus

This unit carries 07 Marks in JKBOSE Board Exam. High-scoring areas: Raoult's Law, Colligative Properties, Henry's Law, Van't Hoff Factor. Master the formulas + one numerical = guaranteed full marks!

Introduction to Solutions

Think of making chai (tea) — you add tea leaves and sugar to water. The water is the solvent (present in largest amount), while tea and sugar are solutes (the dissolved substances). The final chai is a solution — a homogeneous mixture where everything is uniformly mixed at the molecular level.

Formally, a solution is a homogeneous mixture of two or more substances whose composition can be varied within certain limits. A binary solution has exactly two components — one solute and one solvent. This chapter focuses mainly on binary solutions.

Solution vs Suspension vs Colloid

Property	Solution	Colloid	Suspension
Particle size	< 1 nm	1 – 1000 nm	> 1000 nm
Appearance	Transparent	Translucent	Opaque
Tyndall Effect	No	Yes	Yes
Settles on standing	No	No	Yes
Filterable	No	No (passes filter)	Yes
Example	Salt in water	Milk, fog	Muddy water

Fig 1.1: A homogeneous solution — solute particles are evenly distributed among solvent particles

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Types of Solutions

Solutions are classified based on the physical state of the solute and solvent. Since matter exists in three states (solid, liquid, gas), there are nine possible types of binary solutions:

Solute	Solvent	Type	Examples
Gas	Gas	Gaseous	Air (O₂ + N₂); LPG mixture
Liquid	Gas	Gaseous	Water vapour in air; chloroform in N₂
Solid	Gas	Gaseous	Camphor vapour in N₂; smoke particles in air
Gas	Liquid	Liquid	CO₂ in water (soda); O₂ in water
Liquid	Liquid	Liquid	Ethanol in water; antifreeze (glycol + water)
Solid	Liquid	Liquid	Sugar in water; NaCl in water
Gas	Solid	Solid	H₂ in palladium; dissolved gases in minerals
Liquid	Solid	Solid	Amalgam (Hg in Na); moisture in wood
Solid	Solid	Solid	Alloys — Brass (Cu+Zn), Bronze (Cu+Sn)

📌 Remember

The component present in larger quantity is the solvent. Exception: If one component is a liquid, it is usually taken as the solvent regardless of amount.

Types Based on Amount of Solute

Unsaturated Solution: Contains less solute than the maximum it can dissolve at that temperature. More solute can be added.
Saturated Solution: Contains the maximum amount of solute that can dissolve at a given temperature. A dynamic equilibrium exists: Solute (undissolved) ⇌ Solute (dissolved).
Supersaturated Solution: Contains more solute than a saturated solution at the same temperature. It is unstable — adding a tiny crystal (seed crystal) causes excess solute to crystallize out suddenly.

💡 Real-Life Example

Supersaturated solution: Honey is a supersaturated solution of sugar in water. That's why sugar crystals sometimes appear at the bottom of old honey jars!

🎯 Board Tip

The 9-types table is a common 2-mark question. Memorize at least one example for each type. Alloys (brass, bronze, steel) are the most-asked solid-in-solid examples.

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Expression of Concentration of Solutions

Concentration tells us how much solute is present in a given amount of solvent or solution. Different methods are used depending on the context.

1. Mass Percentage (w/w)

The mass of solute present per 100 g of solution.

Mass Percentage

Mass % = w_solutew_solution × 100

Units: % (w/w) — widely used in pharmacy & industry

Example: A 10% glucose solution (w/w) means 10 g of glucose is dissolved in 90 g of water, making 100 g of total solution.

2. Volume Percentage (v/v)

The volume of solute per 100 mL of solution. Used for liquid-in-liquid solutions.

Volume Percentage

Volume % = V_soluteV_solution × 100

Example: 35% ethanol (v/v) = 35 mL ethanol in 100 mL solution

3. Parts per Million (ppm)

Used when solute is present in very small amounts, like pollutants in water or air.

Parts Per Million

ppm = mass of componenttotal mass of solution × 10⁶

Example: 10 ppm of fluoride in water = 10 mg fluoride per kg water

4. Mole Fraction (x)

The ratio of moles of one component to the total moles of all components. It is a dimensionless (no units) quantity.

Mole Fraction

x_A = n_An_A + n_B

Important: x_A + x_B = 1 (always!)

Example: 1 mol ethanol + 3 mol water → x_ethanol = 1/4 = 0.25, x_water = 3/4 = 0.75. Check: 0.25 + 0.75 = 1 ✓

💡 Exam Tip

Mole fraction is always between 0 and 1. The sum of mole fractions of all components always equals 1. This is a great way to double-check your calculations!

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5. Molarity (M)

The number of moles of solute per litre of solution. Most commonly used in labs.

Molarity

M = n_soluteV_solution (in Litres)

Unit: mol L⁻¹ or simply M. Example: 2M NaOH = 2 mol NaOH in 1 L solution

6. Molality (m)

The number of moles of solute per kilogram of solvent (not solution!).

Molality

m = n_solutew_solvent (in kg)

Unit: mol kg⁻¹. Example: 1 m glucose = 1 mol glucose (180 g) in 1 kg water

⚠️ Critical Difference — Most Asked!

Molarity depends on volume → changes with temperature (liquids expand/contract).

Molality depends on mass → does NOT change with temperature.

That's why molality is preferred for studying colligative properties. Board exams ALWAYS ask this!

7. Normality (N) — Brief Mention

Number of gram equivalents of solute per litre of solution. N = M × n-factor. (n-factor = basicity for acids, acidity for bases, change in oxidation state for redox reactions).

Derivations of Inter-conversion Formulas

Derivation 1: Molarity (M) to Molality (m)

Let the volume of the solution be 1 Litre (1000 mL).

Step 1: By definition of Molarity, moles of solute (n₂) = M moles.
Step 2: Mass of solute (w₂) = moles × molar mass = M × M₂ grams (where M₂ is the molar mass of the solute).
Step 3: Mass of solution (w_solution) = Volume × density = 1000 × d grams (where d is the density in g/mL).
Step 4: Mass of solvent (w₁) = Mass of solution − Mass of solute = (1000d − M × M₂) grams.
Step 5: Convert mass of solvent to kg: w₁ (in kg) = 1000d − M × M₂1000.
Step 6: By definition of Molality (m):
m = n₂w₁ (in kg) = M1000d − M × M₂1000 = M × 10001000d − M × M₂

Derived Formula: m = M × 10001000d − M × M₂

Derivation 2: Mole Fraction (x) to Molality (m)

Let x₁ and x₂ be the mole fractions of the solvent and solute respectively, and n₁, n₂ be their number of moles.

Step 1: The ratio of mole fractions is:
x₂x₁ = n₂n₁ + n₂n₁n₁ + n₂ = n₂n₁
Step 2: Moles of solvent (n₁) = w₁M₁ (where w₁ is mass of solvent and M₁ is its molar mass).
Step 3: Substituting n₁ in the ratio:
x₂x₁ = n₂w₁M₁ = n₂ × M₁w₁ ⟹ w₁ = n₂ × M₁ × x₁x₂ grams
Step 4: Convert mass of solvent to kg: w₁ (in kg) = w₁1000 = n₂ × M₁ × x₁1000 × x₂.
Step 5: By definition of Molality (m):
m = n₂w₁ (in kg) = n₂n₂ × M₁ × x₁1000 × x₂ = 1000 × x₂x₁ × M₁

Derived Formula: m = 1000 × x₂x₁ × M₁

🔢 Quick Numerical: Molarity to Molality

Question: Convert 0.5 M NaCl solution (density = 1.02 g/mL) to molality. M₂(NaCl) = 58.5 g/mol.

Step 1: Mass of 1 L solution = 1000 × 1.02 = 1020 g

Step 2: Mass of solute in 1 L = 0.5 × 58.5 = 29.25 g

Step 3: Mass of solvent = 1020 − 29.25 = 990.75 g = 0.991 kg

Step 4: Molality = 0.5 / 0.991 = 0.505 mol/kg

Answer: m = 0.505 mol kg⁻¹

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Solubility

Solubility is the maximum amount of a solute that can dissolve in a specified amount of solvent at a given temperature to form a saturated solution. It depends on the nature of solute, solvent, temperature, and pressure.

Solubility of a Solid in a Liquid

The golden rule is "like dissolves like":

Polar solutes (NaCl, sugar, KNO₃) dissolve in polar solvents (water)
Non-polar solutes (naphthalene, grease) dissolve in non-polar solvents (benzene, CCl₄)
Ionic compounds dissolve in water because the hydration energy released compensates for the lattice energy needed to break the crystal.

Effect of Temperature

Dissolving a solid in a liquid can be endothermic (absorbs heat) or exothermic (releases heat):

If dissolution is endothermic (ΔH_sol > 0): Increasing temperature increases solubility. (Le Chatelier's principle — heat favours the forward process.) Most salts like KNO₃, NaNO₃ follow this.
If dissolution is exothermic (ΔH_sol < 0): Increasing temperature decreases solubility. Examples: Ce₂(SO₄)₃, Ca(OH)₂, CaSO₄.

Effect of Pressure

Pressure has negligible effect on solubility of solids in liquids because solids and liquids are nearly incompressible.

Fig 1.2: Solubility vs Temperature — KNO₃ rises steeply, NaCl barely changes, Ce₂(SO₄)₃ decreases

📌 Remember

Most solids: solubility increases with temperature. Key exceptions: Ce₂(SO₄)₃, Ca(OH)₂, and CaSO₄ — their solubility decreases with temperature.

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Solubility of a Gas in a Liquid

Gas solubility depends on nature of gas, temperature, and pressure. Gases which can be easily liquefied (like CO₂, NH₃, HCl) are more soluble in water.

Henry's Law

Statement: "At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution."

Mathematically: the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.

Henry's Law

p = K_H × x

K_H = Henry's Law constant (specific to each gas-solvent pair). Higher K_H → LOWER solubility!

K_H Values for Common Gases (in water at 298 K)

Gas	K_H (kbar)	Solubility
He	144.97	Very Low
N₂	76.48	Low
O₂	34.86	Moderate
CO₂	1.67	High
HCl	0.067	Very High

Applications of Henry's Law

Carbonated drinks: CO₂ is dissolved under high pressure (~3-4 atm). When you open the bottle, pressure drops → gas escapes → fizz!
Scuba diving (The Bends): At depth, high pressure dissolves more N₂ in blood. Rising too fast releases N₂ as bubbles → painful condition. Tanks use He-O₂ mix (He has high K_H, low solubility).
High altitude sickness: At high altitude, lower O₂ pressure → less O₂ dissolves in blood → anoxia (weakness, inability to think clearly).
Aquatic life: Warm water holds less dissolved O₂ → fish struggle in warm rivers/lakes. Thermal pollution is a real environmental concern.
Anaesthesia: Anaesthetic gases (like N₂O) are administered at controlled pressures to regulate their solubility in blood and brain tissue.

Fig 1.3: Henry's Law — linear relationship between partial pressure and mole fraction

⛔ Common Mistake

Students often think higher K_H = more soluble. It's the opposite! Higher K_H means LOWER solubility. Think of it this way: K_H is how much pressure you need to dissolve the gas — more pressure needed = harder to dissolve.

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Vapour Pressure of Liquid Solutions

Vapour pressure is the pressure exerted by the vapour of a liquid when it is in dynamic equilibrium with its liquid phase in a closed container at a given temperature. It measures how easily molecules escape from the liquid surface.

Raoult's Law (Solutions of Two Volatile Liquids)

Statement: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

Raoult's Law — Volatile Components

p₁ = x₁ · p₁° & p₂ = x₂ · p₂°

p° = vapour pressure of pure component; x = mole fraction in liquid phase

By Dalton's Law of Partial Pressures, the total vapour pressure is:

P_total = p₁ + p₂ = x₁p₁° + x₂p₂°

Since x₁ = 1 − x₂, we can write:

Total Vapour Pressure

P_total = p₁° + (p₂° − p₁°) · x₂

This is a LINEAR equation in x₂ — the graph of P_total vs x₂ is a straight line!

Fig 1.4: Vapour Pressure vs Mole Fraction for an ideal solution — total VP is a straight line

ℹ️ Vapour Composition

The mole fraction in vapour phase is different from the liquid phase. Using Dalton's law: y₁ = p₁ / P_total and y₂ = p₂ / P_total. The more volatile component is present in greater proportion in the vapour.

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Raoult's Law for Non-Volatile Solute

When a non-volatile solute (like glucose, urea, NaCl) is added to a solvent, only the solvent contributes to vapour pressure. The solute molecules sit on the surface and block solvent molecules from escaping → vapour pressure decreases.

Raoult's Law — Non-volatile Solute

p = p° · x_solvent

Since x_solvent < 1, p is always less than p° → VP is lowered

Relative Lowering of Vapour Pressure (RLVP)

Derivation (commonly asked for 3-5 marks):

By Raoult's Law: p = p° · x₁ (where x₁ = mole fraction of solvent)
Lowering of VP: Δp = p° − p = p° − p°x₁ = p°(1 − x₁) = p° · x₂
Relative lowering: p° − pp° = x₂
Since x₂ = n₂/(n₁ + n₂), for dilute solutions (n₂ ≪ n₁):
p° − pp° ≈ n₂n₁ = w₂ × M₁w₁ × M₂
Rearranging to find molar mass of solute:

Molar Mass from RLVP

M₂ = w₂ × M₁ × p°w₁ × Δp

w₂ = mass of solute, M₁ = molar mass of solvent, w₁ = mass of solvent

🎯 Board Score Booster

The RLVP derivation is a frequently asked 3-5 Marks question in JKBOSE. Memorize these 5 steps — write them clearly with proper labels to get full marks. Always start by stating Raoult's Law.

📌 Key Points to Remember

RLVP = (p° − p)/p° = x_solute — this is a colligative property
It depends only on the number of solute particles, not their nature
Valid for dilute solutions of non-volatile solutes
RLVP is always a positive fraction less than 1

🔢 Quick Check

Q: The VP of pure water is 23.8 mmHg at 25°C. Find VP of a solution with x_urea = 0.1.

Solution: x_water = 1 − 0.1 = 0.9. By Raoult's Law: p = 23.8 × 0.9 = 21.42 mmHg

VP of solution = 21.42 mmHg

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Ideal and Non-Ideal Solutions

Ideal Solutions

Solutions that obey Raoult's Law over the entire range of concentration at all temperatures. This happens when:

Solute-solvent (A-B) forces ≈ Solute-solute (A-A) forces ≈ Solvent-solvent (B-B) forces
ΔH_mix = 0 (no heat change on mixing)
ΔV_mix = 0 (no volume change on mixing)

Examples: n-hexane + n-heptane, benzene + toluene, bromoethane + chloroethane

Non-Ideal Solutions — Positive Deviation

A-B forces are weaker than A-A and B-B forces. Molecules escape more easily → VP is higher than predicted by Raoult's Law.

ΔH_mix > 0 (endothermic — heat is absorbed)
ΔV_mix > 0 (volume increases slightly)

Examples: Ethanol + Acetone, CS₂ + Acetone, Ethanol + Cyclohexane, Water + Ethanol

Non-Ideal Solutions — Negative Deviation

A-B forces are stronger than A-A and B-B forces (often due to hydrogen bonding). Molecules find it harder to escape → VP is lower than predicted.

ΔH_mix < 0 (exothermic — heat is released)
ΔV_mix < 0 (volume decreases slightly)

Examples: Chloroform + Acetone (H-bonding!), HNO₃ + Water, Acetic acid + Pyridine

Property	Ideal	Positive Deviation	Negative Deviation
Raoult's Law	Obeyed	p_obs > p_calc	p_obs < p_calc
ΔH_mix	0	> 0 (endothermic)	< 0 (exothermic)
ΔV_mix	0	> 0	< 0
A-B forces vs A-A, B-B	Equal	Weaker	Stronger
Examples	Benzene + Toluene	Ethanol + Acetone	CHCl₃ + Acetone

Fig 1.5: VP vs Mole Fraction — Ideal (straight), Positive deviation (curves above), Negative deviation (curves below)

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Azeotropes

Azeotropes are binary liquid mixtures that have the same composition in liquid and vapour phase and boil at a constant temperature. They cannot be separated by fractional distillation because the vapour has the same composition as the liquid.

Minimum Boiling Azeotropes

Formed by solutions showing large positive deviation. At a particular composition, the VP is maximum → boiling point is minimum.

Example: Ethanol-Water azeotrope — 95.4% ethanol + 4.6% water, boils at 78.1°C. This is why we cannot get 100% pure ethanol by distillation alone!

Maximum Boiling Azeotropes

Formed by solutions showing large negative deviation. At a particular composition, the VP is minimum → boiling point is maximum.

Example: HNO₃-Water azeotrope — 68% HNO₃ + 32% water, boils at 120.5°C.

⚠️ Don't Confuse!

Azeotropes boil at a constant temperature just like pure substances, but they are NOT pure compounds — they are mixtures with a fixed boiling composition.

Colligative Properties — Introduction

The word "colligative" comes from the Latin word "colligatus" meaning "bound together". Colligative properties are properties of dilute solutions containing a non-volatile solute that depend only on the number of solute particles — not on their chemical identity or nature.

The Four Colligative Properties

Relative Lowering of Vapour Pressure — (p° − p)/p° = x₂
Elevation of Boiling Point — ΔT_b = K_b · m
Depression of Freezing Point — ΔT_f = K_f · m
Osmotic Pressure — Π = CRT

💡 Why "Number of Particles" Matters

Consider two solutions: 0.1 m glucose vs 0.1 m NaCl. NaCl dissociates into Na⁺ + Cl⁻ (2 particles), so it has double the colligative effect compared to glucose (which doesn't dissociate). That's why we need the van't Hoff factor!

🎯 Most Important for Board Exam

Colligative properties form the core of this unit. Expect at least one 5-mark question on deriving/explaining a colligative property + one numerical on ΔT_b, ΔT_f, or Π.

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Colligative Property 1: Elevation of Boiling Point

The boiling point of a liquid is the temperature at which its vapour pressure equals the atmospheric pressure. Since adding a non-volatile solute lowers the VP, the solution needs to be heated to a higher temperature to reach atmospheric pressure → boiling point increases.

Elevation of Boiling Point

ΔT_b = K_b × m

K_b = Molal Elevation Constant (Ebullioscopic Constant). For water: K_b = 0.52 K kg mol⁻¹

Derivation of Molar Mass from Boiling Point Elevation

We know that the elevation in boiling point (ΔT_b) is directly proportional to the molality (m) of the solution:

ΔT_b = K_b × m — (Equation 1)

By definition, molality (m) is the number of moles of solute dissolved in 1 kg of solvent:

m = (moles of solute) / (mass of solvent in kg)
m = w₂ / M₂w₁ / 1000 = w₂ × 1000M₂ × w₁ — (Equation 2)

Where:

w₂ = mass of solute in grams
M₂ = molar mass of solute in g/mol
w₁ = mass of solvent in grams

Substituting the value of molality (m) from Equation 2 into Equation 1:

ΔT_b = K_b × w₂ × 1000M₂ × w₁

Rearranging this equation to solve for the molar mass of the solute (M₂):

Derived Molar Mass Formula

M₂ = K_b × w₂ × 1000ΔT_b × w₁

This formula is highly used in board numericals to determine solute molecular mass.

Colligative Property 2: Depression of Freezing Point

The freezing point is where solid and liquid phases have the same VP. Adding solute lowers the liquid's VP, so the intersection point shifts to a lower temperature → freezing point decreases.

Depression of Freezing Point

ΔT_f = K_f × m

K_f = Molal Depression Constant (Cryoscopic Constant). For water: K_f = 1.86 K kg mol⁻¹

Derivation of Molar Mass from Freezing Point Depression

We know that the depression in freezing point (ΔT_f) is directly proportional to the molality (m) of the solution:

ΔT_f = K_f × m — (Equation 1)

Substituting the definition of molality, m = (w₂ × 1000) / (M₂ × w₁), into Equation 1:

ΔT_f = K_f × w₂ × 1000M₂ × w₁

Rearranging this equation to solve for the molar mass of the solute (M₂):

Derived Molar Mass Formula

M₂ = K_f × w₂ × 1000ΔT_f × w₁

Allows accurate determination of solute molar mass by measuring freezing point lowering.

Applications

Antifreeze: Ethylene glycol is added to car radiators → lowers freezing point → prevents water from freezing in winter.
De-icing roads: NaCl or CaCl₂ is sprinkled on icy roads → lowers freezing point → melts ice. CaCl₂ is more effective because it gives 3 ions (i=3).

Fig 1.6: VP vs Temperature — showing both elevation of boiling point and depression of freezing point

💡 Memory Trick

"Solutions struggle to boil and freeze!" — Solution boils HIGHER (elevation) and freezes LOWER (depression) compared to pure solvent.

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Colligative Property 3: Osmotic Pressure

Osmosis

Osmosis is the spontaneous flow of solvent molecules through a semipermeable membrane (SPM) from a dilute solution (or pure solvent) to a concentrated solution. The SPM allows only solvent molecules to pass — not solute particles.

Osmotic Pressure (Π)

Osmotic pressure is the minimum external pressure applied to the solution side to prevent osmosis (i.e., to stop the solvent from flowing in). It is given by the van't Hoff equation:

Osmotic Pressure (van't Hoff Equation)

Π = C × R × T = n₂V × R × T

C = molarity (mol/L), R = 0.0821 L atm K⁻¹ mol⁻¹, T = temperature in Kelvin

Derivation of Molar Mass from Osmotic Pressure

According to the van't Hoff equation for dilute solutions, osmotic pressure (Π) is given by:

Π = C × R × T — (Equation 1)

Where C is the molar concentration (molarity) of the solution, which is the number of moles of solute (n₂) dissolved per Litre of solution (V):

C = n₂V — (Equation 2)

The number of moles of solute (n₂) is the mass of solute in grams (w₂) divided by its molar mass (M₂):

n₂ = w₂M₂ — (Equation 3)

Substituting Equation 3 into Equation 2 gives the concentration C as:

C = w₂M₂ × V — (Equation 4)

Now, substituting the value of concentration C from Equation 4 into Equation 1:

Π = w₂M₂ × V × R × T = w₂ × R × TM₂ × V

Rearranging the equation to solve for the molar mass of the solute (M₂):

Derived Molar Mass Formula

M₂ = w₂ × R × TΠ × V

This method is particularly useful for finding the molecular mass of biomolecules.

💡 Why Osmotic Pressure is Best for Macromolecules

Even dilute solutions of polymers/proteins give large, measurable Π values
ΔT_b and ΔT_f are too tiny to measure accurately for large molecules
Measured at room temperature → proteins don't denature

Fig 1.7: Osmosis — solvent flows from dilute to concentrated side; height difference = osmotic pressure

Isotonic, Hypertonic, and Hypotonic Solutions

Type	Osmotic Pressure	Effect on Cell	Example
Isotonic	Same Π	No change — cell stays normal	0.9% NaCl (saline) with blood
Hypertonic	Higher Π (outside)	Water flows OUT → cell shrinks (crenation)	Concentrated salt solution
Hypotonic	Lower Π (outside)	Water flows IN → cell swells & may burst (hemolysis)	Distilled water

Reverse Osmosis

If pressure greater than osmotic pressure is applied to the solution side, solvent flows in reverse direction (from solution to pure solvent through SPM). This is called reverse osmosis (RO). Main application: desalination of seawater and home RO water purifiers.

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Abnormal Molar Mass & Van't Hoff Factor (i)

Colligative properties depend on the number of particles. When a solute dissociates (like NaCl → Na⁺ + Cl⁻), particle count increases. When it associates (like acetic acid forming dimers in benzene), particle count decreases. This causes the experimentally determined molar mass to differ from the actual value — called abnormal molar mass.

Van't Hoff Factor (i)

Van't Hoff Factor

i = Observed colligative propertyCalculated colligative property

Also: i = Normal molar mass / Abnormal (observed) molar mass

Condition	Value of i	Example	Expected i
No change (non-electrolyte)	i = 1	Glucose, Urea, Sucrose	1
Dissociation	i > 1	NaCl → Na⁺ + Cl⁻	2
Dissociation	i > 1	BaCl₂ → Ba²⁺ + 2Cl⁻	3
Dissociation	i > 1	K₃PO₄ → 3K⁺ + PO₄³⁻	4
Association (dimerization)	i < 1	CH₃COOH in benzene	0.5

Derivations of Relations with Degree of Dissociation/Association

Derivation 1: Van't Hoff Factor & Degree of Dissociation (α)

Consider a solute A which dissociates to give n ions in solution: A ⇌ nB

Let the initial number of moles of solute A be 1, and initial moles of B be 0.
Let α be the degree of dissociation (fraction of solute that dissociates).
At equilibrium:
- Moles of undissociated solute A = 1 − α
- Moles of ions B formed = nα
Total moles of particles at equilibrium = Moles of A + Moles of B = (1 − α) + nα = 1 + (n − 1)α
By definition, the van't Hoff factor (i) is:
i = Total moles of particles at equilibriumInitial moles of particles
i = 1 + (n − 1)α1 ⟹ i = 1 + (n − 1)α
Rearranging for α:
i − 1 = (n − 1)α ⟹ α = i − 1n − 1

Derived Relation: α = i − 1n − 1

Derivation 2: Van't Hoff Factor & Degree of Association (α)

Consider n molecules of a solute A associating to form a polymer A_n: nA ⇌ A_n

Let the initial number of moles of solute A be 1, and initial moles of A_n be 0.
Let α be the degree of association (fraction of solute that associates).
At equilibrium:
- Moles of unassociated solute A = 1 − α
- Moles of associated species A_n formed = α / n
Total moles of particles at equilibrium = (1 − α) + (α / n) = 1 − α(1 − 1/n)
By definition, the van't Hoff factor (i) is:
i = Total moles of particles at equilibriumInitial moles of particles
i = 1 − α(1 − 1n)1 ⟹ i = 1 − α(1 − 1n)
Rearranging for α:
1 − i = α(1 − 1n) ⟹ α = 1 − i1 − 1n = n(1 − i)n − 1
For dimerization (e.g., acetic acid in benzene, n = 2):
i = 1 − α(1 − 12) = 1 − α2 ⟹ α = 2(1 − i)

Derived Relation: α = 1 − i1 − 1n

Modified Colligative Property Equations

All colligative property formulas are multiplied by i:

RLVP: (p° − p)/p° = i · x₂
Boiling Point: ΔT_b = i · K_b · m
Freezing Point: ΔT_f = i · K_f · m
Osmotic Pressure: Π = i · CRT

Fig 1.8: Dissociation increases particles (i > 1); Association decreases particles (i < 1)

🎯 Board Score Booster

The formula i = 1 + (n−1)α is the most tested numerical formula in this chapter. Practice calculating i for NaCl (n=2), CaCl₂ (n=3), K₂SO₄ (n=3), Al₂(SO₄)₃ (n=5), and K₄[Fe(CN)₆] (n=5) with different values of α.

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Solved Numerical Problems

Problem 1: Mole Fraction of Benzene

Q. Calculate the mole fraction of benzene in a solution containing 30% by mass of benzene in carbon tetrachloride.

Step 1: Identify masses

Mass of benzene (C₆H₆) = 30 g | Mass of CCl₄ = 100 − 30 = 70 g

Step 2: Calculate molar masses

M(C₆H₆) = 78 g/mol | M(CCl₄) = 154 g/mol

Step 3: Calculate moles

n(benzene) = 30 / 78 = 0.385 mol

n(CCl₄) = 70 / 154 = 0.454 mol

Step 4: Apply mole fraction formula

x_benzene = n_benzene / (n_benzene + n_CCl₄) = 0.385 / (0.385 + 0.454) = 0.385 / 0.839

x_benzene = 0.459

Problem 2: Molarity of Na₂CO₃ Solution

Q. Calculate the molarity of a solution containing 5.3 g of Na₂CO₃ dissolved in 250 mL of solution.

Step 1: Calculate molar mass

M(Na₂CO₃) = 2(23) + 12 + 3(16) = 106 g/mol

Step 2: Calculate moles of solute

n = 5.3 / 106 = 0.05 mol

Step 3: Convert volume to litres

V = 250 / 1000 = 0.25 L

Step 4: Apply molarity formula

Molarity (M) = n / V = 0.05 / 0.25

Molarity = 0.2 M

Problem 3: Vapour Pressure by Raoult's Law

Q. The vapour pressure of pure water at 25°C is 23.8 mm Hg. Find the vapour pressure of an aqueous solution of urea with mole fraction of urea = 0.1.

Step 1: Find mole fraction of solvent

x_water = 1 − x_urea = 1 − 0.1 = 0.9

Step 2: Apply Raoult's Law

p = p° × x_water = 23.8 × 0.9

p = 21.42 mm Hg

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Solved Numerical Problems (Cont.)

Problem 4: Molality from Boiling Point Elevation

Q. A solution of glucose (M = 180 g/mol) in water has a boiling point of 100.20°C. Calculate the molality of the solution. (K_b for water = 0.52 K kg/mol)

Step 1: Calculate ΔT_b

ΔT_b = T_b(solution) − T_b(solvent) = 100.20 − 100 = 0.20°C

Step 2: Apply elevation formula

ΔT_b = K_b × m ⟹ m = ΔT_b / K_b = 0.20 / 0.52

m = 0.385 mol/kg

Problem 5: Molality & Van't Hoff Factor for NaCl

Q. An aqueous solution of NaCl freezes at −0.372°C. What is the molality? (K_f = 1.86 K kg/mol). Also calculate the van't Hoff factor if the expected ΔT_f for 0.1 m NaCl is 0.186°C but observed is 0.372°C.

Step 1: Calculate molality

ΔT_f = 0.372°C

m = ΔT_f / K_f = 0.372 / 1.86 = 0.2 mol/kg

Step 2: Calculate van't Hoff factor

i = Observed ΔT_f / Calculated ΔT_f = 0.372 / 0.186 = 2

Step 3: Verify logically

NaCl → Na⁺ + Cl⁻ (2 particles per formula unit), so i = 2 ✓

m = 0.2 mol/kg | i = 2

Problem 6: Osmotic Pressure of K₂SO₄ Solution

Q. Calculate the osmotic pressure of a solution obtained by dissolving 1.95 g of K₂SO₄ in 250 mL of solution at 27°C. Assume complete dissociation. (R = 0.0821 L atm mol⁻¹ K⁻¹, M(K₂SO₄) = 174)

Step 1: Calculate moles

n = 1.95 / 174 = 0.01121 mol

Step 2: Calculate molar concentration

C = 0.01121 / 0.25 = 0.04484 M

Step 3: Find van't Hoff factor

K₂SO₄ → 2K⁺ + SO₄²⁻ ⟹ i = 3

Step 4: Convert temperature

T = 27 + 273 = 300 K

Step 5: Apply Π = iCRT

Π = 3 × 0.04484 × 0.0821 × 300

Π = 3.31 atm

🎯 Exam Strategy for Numericals

In numerical questions, always:

(1) Write the formula first
(2) Convert all units properly (mL → L, g → kg, °C → K)
(3) Don't forget the van't Hoff factor (i) for electrolytes!

Asterisk Classes

Important PYQs with Model Answers (JKBOSE)

1. State Henry's Law and mention two applications. JKBOSE 2023

Henry's Law: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution.

Mathematical form: p = K_H × x

Applications:

(i) Carbonated beverages — CO₂ is dissolved under high pressure. When the bottle is opened, pressure drops → gas escapes as bubbles.
(ii) Scuba diving — Air is diluted with helium to prevent nitrogen narcosis ("the bends") at high underwater pressures.

2. Define molality. Why is it preferred over molarity? JKBOSE 2023

Molality (m) is the number of moles of solute per kilogram of solvent.

It is preferred over molarity because it is independent of temperature (based on mass of solvent), unlike molarity which changes with temperature (based on volume of solution — volume expands/contracts with temperature).

3. What are colligative properties? Name all four. JKBOSE 2022

Colligative properties are properties of dilute solutions that depend only on the number of solute particles, not on their chemical identity.

The four colligative properties are:

(i) Relative lowering of vapour pressure
(ii) Elevation of boiling point
(iii) Depression of freezing point
(iv) Osmotic pressure

4. What is van't Hoff factor? What is its value for association? JKBOSE 2022

Van't Hoff factor (i) = Observed colligative property / Calculated colligative property

For association: particles combine → number of particles decreases → i < 1.

Example: Acetic acid in benzene forms dimers → i ≈ 0.5.

5. Differentiate ideal and non-ideal solutions. JKBOSE 2021

Property	Ideal Solutions	Non-Ideal (+ deviation)	Non-Ideal (− deviation)
Raoult's Law	Obeyed	p_obs > p_Raoult	p_obs < p_Raoult
ΔH_mix	0	> 0 (endothermic)	< 0 (exothermic)
ΔV_mix	0	> 0	< 0
A–B forces vs A–A, B–B	Equal	Weaker	Stronger
Example	Benzene + Toluene	Ethanol + Acetone	Chloroform + Acetone

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Important PYQs with Model Answers (Cont.)

6. What is reverse osmosis? Give its application. JKBOSE 2021

When external pressure greater than osmotic pressure is applied on the solution side, the solvent flows from the solution to the pure solvent through a semi-permeable membrane. This is called reverse osmosis.

Application: Desalination of seawater to obtain fresh drinking water.

7. Why does a solution have a higher boiling point than the pure solvent? JKBOSE 2020

A non-volatile solute lowers the vapour pressure of the solution. Since the vapour pressure is lower, a higher temperature is needed for the VP to become equal to the atmospheric pressure → hence, the boiling point is elevated.

8. Calculate the mole fraction of ethylene glycol in a 20% solution by mass. JKBOSE 2020

In 100 g of solution: Mass of glycol = 20 g, Mass of water = 80 g.

M(glycol, C₂H₆O₂) = 62 g/mol | M(H₂O) = 18 g/mol

n(glycol) = 20/62 = 0.322 mol

n(water) = 80/18 = 4.444 mol

x_glycol = 0.322 / (0.322 + 4.444) = 0.322 / 4.766 = 0.068

9. What are azeotropes? Give an example of a minimum boiling azeotrope. JKBOSE 2019

Azeotropes are binary liquid mixtures that boil at a constant temperature and have the same composition in both the liquid and vapour phases. They cannot be separated by simple distillation.

Minimum boiling azeotrope: Ethanol–water mixture (95.4% ethanol, bp 78.1°C). This is formed by solutions showing positive deviation from Raoult's law.

10. Define osmotic pressure. Why is it preferred for determining molar mass of macromolecules? JKBOSE 2019

Osmotic pressure (Π) is the minimum pressure applied on the solution side to prevent the flow of solvent through a semi-permeable membrane into the solution.

Preferred for macromolecules because:

(i) Even dilute solutions produce a large, measurable osmotic pressure value.
(ii) Measurement is done at room temperature — this prevents denaturation of proteins and biomolecules.

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Important PYQs with Model Answers (Cont.)

11. What is the effect of temperature on the solubility of a gas in a liquid? JKBOSE 2018

The solubility of a gas in a liquid decreases with increasing temperature.

Reason: Dissolution of gas is an exothermic process. By Le Chatelier's principle, increasing temperature shifts the equilibrium backward (towards gas release) → solubility decreases.

12. What are isotonic solutions? JKBOSE 2018

Isotonic solutions are solutions that have the same osmotic pressure at the same temperature. There is no net flow of solvent between them when separated by a semi-permeable membrane.

Example: 0.9% NaCl solution (normal saline) is isotonic with blood — hence used in IV drips.

13. State Raoult's law for a solution containing a non-volatile solute. JKBOSE 2017

Raoult's Law: The vapour pressure of a solution containing a non-volatile solute equals the product of the vapour pressure of the pure solvent and the mole fraction of the solvent.

Mathematically: p = p° × x_solvent

14. Why is the vapour pressure of a glucose solution lower than that of pure water? JKBOSE 2017

Glucose molecules occupy surface area at the liquid–gas interface. This means fewer water molecules are available at the surface to escape into the vapour phase → the rate of evaporation decreases → vapour pressure is lowered.

15. What is positive deviation from Raoult's law? JKBOSE 2016

When the observed vapour pressure of a solution is greater than that predicted by Raoult's law:

p_observed > p_Raoult

Cause: A–B intermolecular forces are weaker than A–A and B–B forces.

ΔH_mix > 0, ΔV_mix > 0

Example: Ethanol + Acetone

16. Why is K_H higher for gases with lower solubility? JKBOSE 2016

From Henry's Law: p = K_H × x

For a given partial pressure (p), if K_H is large, then x (mole fraction of gas) must be small → low solubility. Hence, higher K_H = lower solubility.

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Important PYQs with Model Answers (Cont.)

17. Define ebullioscopic constant (K_b). JKBOSE 2015

The ebullioscopic constant (K_b) is the elevation in boiling point when the molality of the solution is 1 (i.e., 1 mole of solute dissolved per kg of solvent).

Unit: K kg mol⁻¹ (or °C kg mol⁻¹)

18. What is abnormal molecular mass? JKBOSE 2015

Abnormal molecular mass is the molar mass determined from colligative property measurements that differs from the theoretically expected value.

Cause: Association (observed M > theoretical M, i < 1) or dissociation (observed M < theoretical M, i > 1) of solute particles in solution.

19. Why is salt sprinkled on snowy roads in winter? JKBOSE 2014

When salt (NaCl) is sprinkled on snow, it dissolves in the thin water film and causes a depression in the freezing point. The freezing point drops below 0°C, causing the ice to melt even at sub-zero temperatures → roads become safe for driving.

20. K₄[Fe(CN)₆] is 50% dissociated in solution. Find the van't Hoff factor (i). JKBOSE 2014

Dissociation: K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻

Number of ions (n) = 4 + 1 = 5

Degree of dissociation: α = 50% = 0.5

Formula: i = 1 + (n − 1)α = 1 + (5 − 1)(0.5) = 1 + 4 × 0.5 = 1 + 2

i = 3

🔁 PYQ Pattern Observation

Questions on Henry's Law, colligative properties, Raoult's law, and van't Hoff factor appear almost every year. Master these topics thoroughly for a guaranteed 5–7 marks!

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Additional Important Questions

21. What is the relationship between the van't Hoff factor and degree of dissociation? Important

For a solute that dissociates into n ions:

i = 1 + (n − 1)α

where α = degree of dissociation.

Rearranging: α = (i − 1) / (n − 1)

22. What happens when a plant cell is placed in a hypertonic solution?

Water moves out of the cell through osmosis → the cell membrane shrinks away from the cell wall → this phenomenon is called plasmolysis. The cell becomes flaccid.

23. Derive the relationship for relative lowering of vapour pressure. 5 Marks

By Raoult's law for a non-volatile solute: p = p°x₁

Since x₁ + x₂ = 1, we have x₁ = 1 − x₂

∴ p = p°(1 − x₂)

∴ p° − p = p°x₂

∴ p° − pp° = x₂ = n₂n₁ + n₂

For dilute solutions, n₂ ≪ n₁:

p° − pp° ≈ n₂n₁ = w₂ / M₂w₁ / M₁

Rearranging to find molar mass:

M₂ = w₂ × M₁ × p°w₁ × (p° − p)

24. Why is the molecular mass of acetic acid found to be about 120 when dissolved in benzene, although its actual molecular mass is 60?

Acetic acid forms dimers in benzene through hydrogen bonding:

2 CH₃COOH ⇌ (CH₃COOH)₂

Two molecules associate into one particle → the number of particles is halved → colligative properties are lower than expected → the calculated molar mass is roughly double the actual value (120 instead of 60).

Here, i ≈ 0.5 (i < 1, indicating association).

25. Calculate the molality of 2.5 g of ethanoic acid (CH₃COOH) dissolved in 75 g of benzene.

M(CH₃COOH) = 12 + 3(1) + 12 + 16 + 16 + 1 = 60 g/mol

Moles of solute = 2.5 / 60 = 0.0417 mol

Mass of solvent = 75 / 1000 = 0.075 kg

Molality = 0.0417 / 0.075 = 0.556 m

Asterisk Classes

Additional Important Questions (Cont.)

26. Explain why cooking is difficult at hill stations using the concept of colligative properties.

At high altitude, atmospheric pressure is lower. Water boils when its vapour pressure equals atmospheric pressure. Since atmospheric pressure is reduced, water boils at a temperature lower than 100°C → food takes longer to cook.

Solution: A pressure cooker is used — it increases the internal pressure, which raises the boiling point of water → food cooks faster.

27. What is the effect of adding a non-volatile solute on the freezing point of a solvent? Explain with a VP–temperature diagram. 3 Marks

Adding a non-volatile solute lowers the vapour pressure of the solution.

The freezing point is the temperature at which the VP of the liquid phase = VP of the solid phase.

Since VP of the solution curve lies below the VP curve of the pure solvent, the intersection with the solid VP curve shifts to a lower temperature.

∴ Depression in freezing point: ΔT_f = T_f° − T_f(solution)

Fig: VP–Temperature diagram showing depression in freezing point

28. Two elements A and B form compounds AB₂ and AB₄. When dissolved in 20 g of benzene, 1 g of AB₂ lowers the freezing point by 2.3 K while 1 g of AB₄ lowers it by 1.3 K. K_f for benzene = 5.1 K kg/mol. Find atomic masses of A and B.

For AB₂: M = K_f × w₂ × 1000 / (ΔT_f × w₁)

M = 5.1 × 1 × 1000 / (2.3 × 20) = 5100 / 46 ≈ 111 g/mol

For AB₄: M = 5.1 × 1 × 1000 / (1.3 × 20) = 5100 / 26 ≈ 196 g/mol

Let atomic mass of A = a, B = b:

a + 2b = 111 …(i)

a + 4b = 196 …(ii)

Subtracting (i) from (ii): 2b = 85 → b = 42.5

From (i): a = 111 − 2(42.5) = 111 − 85 = a = 26

29. Why is the osmotic pressure method preferred for determining the molar mass of polymers and proteins?

(i) Osmotic pressure generates large, measurable values even for very dilute solutions of macromolecules.
(ii) Other methods (ΔT_b, ΔT_f) give very small changes for dilute solutions — difficult to measure accurately.
(iii) Osmotic pressure is measured at room temperature, preventing denaturation of biomolecules.

30. State Raoult's law for solutions of volatile liquids. What is the composition of vapour in equilibrium with the solution? 5 Marks

Raoult's Law: The partial vapour pressure of each volatile component is equal to the product of its mole fraction in the solution and its vapour pressure in the pure state.

p₁ = x₁ p₁° & p₂ = x₂ p₂°

Total vapour pressure (Dalton's law): P_total = p₁ + p₂ = x₁p₁° + x₂p₂°

Composition of vapour:

y₁ = p₁ / P_total & y₂ = p₂ / P_total

where y₁ and y₂ are the mole fractions in the vapour phase.

Asterisk Classes

⚠️ Common Mistakes to Avoid

❌ Mistake 1: Confusing Molarity and Molality

Molarity (M) = moles of solute / litres of SOLUTION

Molality (m) = moles of solute / kilograms of SOLVENT

Remember: molaRity → R for "peR litRe of solution." molaLity → L for "per kiLo of soLvent."

❌ Mistake 2: Forgetting Van't Hoff Factor

For electrolytes (NaCl, BaCl₂, K₂SO₄, etc.), you MUST multiply the colligative property by i.

ΔT_b = i × K_b × m | Π = i × CRT

Forgetting 'i' = losing marks on an easy question!

❌ Mistake 3: Sign of ΔT

ΔT_b is always positive (boiling point goes UP).

ΔT_f is also always positive (it represents the magnitude of depression).

T_f(solution) = T_f(solvent) − ΔT_f ← subtract, don't add!

❌ Mistake 4: Henry's Law Constant Confusion

Higher K_H means LESS soluble, NOT more soluble!

p = K_H × x → Large K_H → small x at given p → low solubility.

❌ Mistake 5: Mixing Up Positive & Negative Deviations

Positive deviation: Weaker A–B forces → molecules escape more easily → higher VP

Negative deviation: Stronger A–B forces → molecules held tightly → lower VP

❌ Mistake 6: Misapplying Raoult's Law

Raoult's Law applies strictly to ideal solutions or as a limiting law for very dilute solutions. Don't apply it blindly to non-ideal solutions without accounting for deviations.

❌ Mistake 7: Unit Errors

K_b is in K kg mol⁻¹, NOT K/mol.

Molality is in mol/kg, NOT mol/g.

Always convert: mL → L (÷1000), g → kg (÷1000), °C → K (+273).

❌ Mistake 8: Osmotic Pressure Direction

Solvent flows from DILUTE to CONCENTRATED (low solute concentration → high solute concentration).

NOT the other way around! Think: solvent moves to "dilute" the more concentrated side.

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📋 Quick Revision — All Formulas at a Glance

Property / Concept	Formula	Units	Notes
Mass Percentage	Mass % = w_solutew_solution × 100	%	w/w basis
Volume Percentage	Vol % = V_soluteV_solution × 100	%	v/v basis
Parts per Million	ppm = mass of componenttotal mass × 10⁶	ppm	For very dilute solutions
Mole Fraction	x_A = n_An_A + n_B	Unitless	x_A + x_B = 1
Molarity	M = n_soluteV_solution (L)	mol L⁻¹	Temperature dependent
Molality	m = n_solutew_solvent (kg)	mol kg⁻¹	Temperature independent
Henry's Law	p = K_H × x	K_H: atm or bar	Higher K_H → lower solubility
Raoult's Law (volatile)	p_i = x_i × p_i°	mm Hg / atm	For each component
Raoult's Law (RLVP)	(p° − p) / p° = x₂	Unitless	x₂ = mole fraction of solute
Elevation in B.P.	ΔT_b = K_b × m	K or °C	K_b: K kg mol⁻¹
Depression in F.P.	ΔT_f = K_f × m	K or °C	K_f: K kg mol⁻¹
Osmotic Pressure	Π = CRT	atm	C in mol/L, T in K
With van't Hoff factor	Multiply each by i	—	ΔT_b = iK_bm, Π = iCRT
Dissociation	i = 1 + (n − 1)α	Unitless	n = number of ions; i > 1
Association	i = 1 − (1 − 1n)α	Unitless	n = molecules associating; i < 1
M₂ from B.P. elevation	M₂ = K_b × w₂ × 1000ΔT_b × w₁	g/mol	w₂, w₁ in grams
M₂ from F.P. depression	M₂ = K_f × w₂ × 1000ΔT_f × w₁	g/mol	w₂, w₁ in grams
M₂ from Osmotic Pressure	M₂ = w₂ × R × TΠ × V	g/mol	V in litres, T in K

💡 Quick Memory Hack

The three "M₂ formulas" have the same structure: M₂ = (constant × w₂ × factor) / (observed change × solvent/volume). Learn one, derive the rest!

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🎯 Mark-wise Question Bank for Practice

Section A: 1-Mark Questions

What is a binary solution?
Define molality.
State Henry's law.
What is osmotic pressure?
Name the colligative property used in determining molecular mass by Landsberger's method.
What type of deviation is shown by ethanol–water mixture?
Define cryoscopic constant.
What is the van't Hoff factor for glucose in water?

Section B: 2-Mark Questions

Distinguish between molarity and molality.
Why is the boiling point of 1 M NaCl higher than 1 M glucose?
What are isotonic solutions? Give an example.
Why is osmotic pressure preferred for determining molar mass of macromolecules?
Calculate the molality of a 10% NaOH solution (w/w).
Why does molarity change with temperature?

Section C: 3-Mark Questions

Derive the formula for relative lowering of vapour pressure.
Explain Henry's Law with three applications.
Differentiate between ideal and non-ideal solutions with examples.
Explain the depression in freezing point of a solvent on addition of a non-volatile solute.
Calculate the boiling point of a solution of 1.8 g glucose in 100 g water. (K_b = 0.52 K kg mol⁻¹)

Section D: 5-Mark Questions

What are colligative properties? Explain elevation in boiling point and derive its formula.
What is osmosis? Define osmotic pressure. Explain isotonic, hypertonic, and hypotonic solutions with examples.
Define Raoult's law. Explain positive and negative deviations with graphs and examples.
What is van't Hoff factor? How is it related to degree of dissociation? Solve: 0.1 M BaCl₂ is 80% dissociated. Find i.

🌟 All the Best!

All the best for your JKBOSE Board Exam! Remember: Consistent revision + Practicing numericals = Guaranteed high marks.

— Balvir Sir, Asterisk Classes

Unit-I: Solutions JKBOSE Class 12 Chemistry — Complete Study Material

Introduction to Solutions

Solution vs Suspension vs Colloid

Types of Solutions

Types Based on Amount of Solute

Expression of Concentration of Solutions

1. Mass Percentage (w/w)

2. Volume Percentage (v/v)

3. Parts per Million (ppm)

4. Mole Fraction (x)

5. Molarity (M)

6. Molality (m)

7. Normality (N) — Brief Mention

Derivations of Inter-conversion Formulas

Solubility

Solubility of a Solid in a Liquid

Effect of Temperature

Effect of Pressure

Solubility of a Gas in a Liquid

Henry's Law

KH Values for Common Gases (in water at 298 K)

Applications of Henry's Law

Vapour Pressure of Liquid Solutions

Raoult's Law (Solutions of Two Volatile Liquids)

Raoult's Law for Non-Volatile Solute

Relative Lowering of Vapour Pressure (RLVP)

Ideal and Non-Ideal Solutions

Ideal Solutions

Non-Ideal Solutions — Positive Deviation

Non-Ideal Solutions — Negative Deviation

Azeotropes

Minimum Boiling Azeotropes

Maximum Boiling Azeotropes

Colligative Properties — Introduction

The Four Colligative Properties

Colligative Property 1: Elevation of Boiling Point

Derivation of Molar Mass from Boiling Point Elevation

Colligative Property 2: Depression of Freezing Point

Derivation of Molar Mass from Freezing Point Depression

Applications

Colligative Property 3: Osmotic Pressure

Osmosis

Osmotic Pressure (Π)

Derivation of Molar Mass from Osmotic Pressure

Isotonic, Hypertonic, and Hypotonic Solutions

Reverse Osmosis

Abnormal Molar Mass & Van't Hoff Factor (i)

Van't Hoff Factor (i)

Derivations of Relations with Degree of Dissociation/Association

Modified Colligative Property Equations

Solved Numerical Problems

Solved Numerical Problems (Cont.)

Important PYQs with Model Answers (JKBOSE)

Important PYQs with Model Answers (Cont.)

Important PYQs with Model Answers (Cont.)

Important PYQs with Model Answers (Cont.)

Additional Important Questions

Additional Important Questions (Cont.)

⚠️ Common Mistakes to Avoid

📋 Quick Revision — All Formulas at a Glance

🎯 Mark-wise Question Bank for Practice

Section A: 1-Mark Questions

Section B: 2-Mark Questions

Section C: 3-Mark Questions

Section D: 5-Mark Questions

K_H Values for Common Gases (in water at 298 K)