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🎓 Jnaanangkur – The Learning Hub  |  Class 12 Chemistry  |  Chapter 1: Solutions  |  CBSE & State Board 2024–25

⚗️

📚 Complete Study Guide

Class 12 Chemistry
Chapter 1: Solutions

NCERT-based | Board Exam Ready | MCQs + Numericals + PYQs + Mind Maps | CBSE & State Board

✅ NCERT Aligned 📊 50+ MCQs 🔢 20+ Numericals 📝 PYQ 2018–2024 🧠 HOTS Questions ⚡ Formula Sheet

📋 Table of Contents

🎓 Welcome Note 📖 Chapter Overview 📝 Key Concepts ⭐ Important Definitions ⚡ Formula Sheet 💡 Memory Tricks 📊 Concept Maps ❓ NCERT Solutions 📚 PYQ Questions 🎯 Important Questions 🧮 Numericals 🧠 HOTS Questions 🚫 Common Mistakes 📋 MCQ Practice 🚀 Quick Revision 🎯 Exam Tips

🎓

Welcome Note

A message for every aspiring Class 12 student

Dear Class 12 Student — You've Got This! 🌟

Welcome to the most important year of your school journey! Chapter 1 — Solutions — is one of the highest-scoring chapters in Class 12 Chemistry. Every year, 8–12 marks are directly asked from this chapter in CBSE Board exams, and it forms a critical base for JEE, NEET, and all State Board competitive exams.

From the dissolved sugar in your morning tea ☕ to the saline drip in a hospital 🏥, solutions are everywhere around you. Understanding this chapter means understanding chemistry as it exists in your daily life.

This guide is your complete companion — every formula, every concept, every important question is right here. Read it, revise it, and conquer your board exams with full confidence! 💪

📖

Chapter Overview

Introduction and real-life applications

🔬 What is a Solution?

A solution is a homogeneous mixture of two or more substances, where the composition is uniform throughout. The substance present in larger amount is called the solvent, and the one in smaller amount is called the solute.

• Solutions can be solid, liquid, or gaseous
• They have variable composition
• They do not scatter light (unlike colloids)
• Cannot be separated by filtration

🌍 Real-Life Applications

• 🧂 Salt in water — cooking
• 💉 Saline (0.9% NaCl) — medical drips
• 🩸 Blood plasma — biological solution
• 🌊 Seawater — aqueous salt solution
• 🚗 Antifreeze (ethylene glycol) — car radiators
• 🍷 Alcohol in water — beverages
• 🌡️ Thermometers — mercury (liquid in liquid)
• 💧 Aerated drinks — CO₂ in water (gas in liquid)

💡

Did You Know?

The human blood plasma is essentially an aqueous solution containing proteins, salts, glucose, and hormones. Its osmotic pressure must be carefully maintained — this is why hospitals use isotonic saline rather than pure water in IV drips!

📋 Types of Solutions (9 Types)

State of Solute	State of Solvent	Type	Example
Gas	Gas	Gaseous solution	Air (O₂, N₂ in N₂)
Liquid	Gas	Gaseous solution	Chloroform in N₂
Solid	Gas	Gaseous solution	Camphor in N₂
Gas	Liquid	Liquid solution	CO₂ in water (soda)
Liquid	Liquid	Liquid solution	Ethanol in water
Solid	Liquid	Liquid solution	NaCl in water (most common)
Gas	Solid	Solid solution	H₂ in palladium
Liquid	Solid	Solid solution	Amalgam (Hg in Ag)
Solid	Solid	Solid solution	Alloys (Cu in Au)

📝

Teacher's Key Concepts

All important topics explained clearly

📏 Concentration of Solutions

Concentration tells us how much solute is dissolved in a given amount of solvent/solution. There are multiple ways to express it:

Term	Formula	Units	Temperature Dependence
Mass %	(mass of solute / mass of solution) × 100	%	Independent
Volume %	(volume of solute / volume of solution) × 100	%	Independent
Mole Fraction (χ)	nA / (nA + nB)	Dimensionless	Independent
Molarity (M)	moles of solute / volume of solution (L)	mol L⁻¹	Dependent ⚠️
Molality (m)	moles of solute / mass of solvent (kg)	mol kg⁻¹	Independent ✅
Normality (N)	gram equivalents / volume of solution (L)	N or eq/L	Dependent ⚠️
ppm	(mass of component / mass of solution) × 10⁶	mg/kg	Independent

💧 Solubility

Solubility is the maximum amount of solute that can dissolve in a specified amount of solvent at a given temperature to produce a saturated solution.

🌡️ Effect of Temperature

• Endothermic dissolution: solubility increases with temperature (NaCl, KNO₃)
• Exothermic dissolution: solubility decreases with temperature (Ce₂(SO₄)₃)

🔄 Henry's Law (Gas in Liquid)

• p = K_H × χ
• At constant temperature, solubility of gas ∝ partial pressure
• K_H is Henry's law constant
• Higher K_H = lower solubility

⚠️ Applications of Henry's Law

• Carbonation of soft drinks (CO₂ at high pressure)
• Scuba divers — nitrogen narcosis (N₂ dissolves at high pressure)
• Anaesthesia using N₂O
• Oxygen cylinders for patients with respiratory disorders

💨 Vapour Pressure & Raoult's Law

Vapour pressure of a liquid is the pressure exerted by its vapour in equilibrium with the liquid at a given temperature. When a non-volatile solute is dissolved, vapour pressure of the solution decreases.

📌 Raoult's Law

• For volatile solute: p_A = χ_A × p°_A  and  p_B = χ_B × p°_B
• Total pressure: P = p_A + p_B = χ_Ap°_A + χ_Bp°_B
• For non-volatile solute: p_s = χ_A × p°_A
• Relative lowering: (p°_A − p_s) / p°_A = χ_B

✅ Ideal Solutions

• Obey Raoult's Law over entire range
• ΔH_mix = 0
• ΔV_mix = 0
• A–A, B–B, A–B interactions are equal
• Examples: Benzene–Toluene, n-hexane–n-heptane, Ethyl bromide–Ethyl iodide

❌ Non-Ideal Solutions

• +ve deviation: p > p_ideal, A–B < A–A/B–B, ΔH_mix > 0
  e.g., Ethanol + Water, Acetone + CS₂
• −ve deviation: p < p_ideal, A–B > A–A/B–B, ΔH_mix < 0
  e.g., Chloroform + Acetone, HNO₃ + Water

🔢 Colligative Properties

💡 Key Concept

Colligative properties depend only on the number of solute particles, NOT on their nature. There are four colligative properties:

1️⃣ Relative Lowering of Vapour Pressure (RLVP) ▼

Formula: Δp/p°_A = χ_B = n_B / (n_A + n_B)

When a non-volatile solute is added, the vapour pressure of the solvent decreases. This decrease, relative to the vapour pressure of pure solvent, is called RLVP.

For dilute solutions: Δp/p° = n_B/n_A = w_BM_A / (M_Bw_A)

2️⃣ Elevation of Boiling Point (ΔT_b) ▼

Formula: ΔT_b = K_b × m

When a non-volatile solute is added, the boiling point of the solution increases. The solution boils at a higher temperature than the pure solvent.

• K_b = Ebullioscopic constant (mol. elevation constant)
• m = molality of solution
• K_b for water = 0.52 K kg mol⁻¹

3️⃣ Depression of Freezing Point (ΔT_f) ▼

Formula: ΔT_f = K_f × m

When a non-volatile solute is added, the freezing point of the solution decreases. The solution freezes at a lower temperature.

• K_f = Cryoscopic constant (mol. depression constant)
• K_f for water = 1.86 K kg mol⁻¹
• Application: Antifreeze in car radiators, salting roads in winter

4️⃣ Osmotic Pressure (π) ▼

Formula: π = CRT = (n/V)RT

Osmosis is the spontaneous flow of solvent molecules through a semi-permeable membrane from a region of lower concentration to higher concentration. The pressure needed to stop this flow is osmotic pressure.

✅ Isotonic Solutions

π₁ = π₂ (no osmosis occurs)

Blood is isotonic with 0.9% NaCl

🔵 Uses of Osmotic Pressure

• Determine molar mass of polymers, proteins
• Reverse osmosis (water purification)
• Medical applications

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

⭐ Van't Hoff Factor

i = (Observed colligative property) / (Calculated colligative property)

i = (Normal molar mass) / (Abnormal molar mass)

📈 Association (i < 1)

• Particles associate → fewer particles
• Observed molar mass > Normal
• Example: Benzoic acid in benzene (dimerizes)
• Acetic acid in benzene

📉 Dissociation (i > 1)

• Particles dissociate → more particles
• Observed molar mass < Normal
• Example: KCl → K⁺ + Cl⁻ (i = 2)
• K₂SO₄ → 2K⁺ + SO₄²⁻ (i = 3)

📌 Modified Colligative Property Formulas with Van't Hoff Factor

• ΔT_b = i × K_b × m
• ΔT_f = i × K_f × m
• π = i × CRT
• Δp/p° = i × χ_B

⭐

Important Definitions

All key terms for quick revision

Solution: A homogeneous mixture of two or more components having uniform composition and properties throughout.

Solute: The component present in smaller amount in the solution.

Solvent: The component present in larger amount; determines the physical state of the solution.

Mole Fraction: The ratio of the number of moles of a component to the total number of moles of all components in the solution. Sum of all mole fractions = 1.

Molarity (M): Number of moles of solute dissolved per litre of solution. Unit: mol L⁻¹. Temperature dependent.

Molality (m): Number of moles of solute dissolved per kilogram of solvent. Unit: mol kg⁻¹. Temperature independent (preferred for colligative properties).

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 liquid. p = K_H × χ.

Raoult's Law: The partial vapour pressure of each volatile component of a solution is directly proportional to its mole fraction. p_i = χ_i × p°_i.

Ideal Solution: A solution that obeys Raoult's Law over the entire range of concentration. ΔH_mix = 0 and ΔV_mix = 0.

Colligative Properties: Properties of solutions that depend only on the number of solute particles, not their nature. Four types: RLVP, elevation of BP, depression of FP, osmotic pressure.

Osmosis: The spontaneous flow of solvent molecules through a semi-permeable membrane from a dilute solution (or pure solvent) to a concentrated solution.

Osmotic Pressure (π): The minimum excess pressure that must be applied on the solution side to prevent osmosis. π = CRT.

Isotonic Solutions: Solutions having equal osmotic pressure. No net osmosis occurs between them.

Van't Hoff Factor (i): Ratio of observed colligative property to calculated colligative property; accounts for solute dissociation or association.

Ebullioscopic Constant (K_b): The elevation in boiling point when 1 mole of non-volatile solute is dissolved in 1 kg of solvent. For water: 0.52 K kg mol⁻¹.

Cryoscopic Constant (K_f): The depression in freezing point when 1 mole of non-volatile solute is dissolved in 1 kg of solvent. For water: 1.86 K kg mol⁻¹.

⚡

Complete Formula Sheet

All formulas with units and explanations

Mole Fraction

χ_A = n_A / (n_A + n_B)

Dimensionless | Range: 0 to 1 | χ_A + χ_B = 1

n = moles of component

Molarity (M)

M = (w_B × 1000) / (M_B × V_mL)

Unit: mol L⁻¹ or M | Temperature dependent

w_B = mass of solute (g), M_B = molar mass, V = volume (mL)

Molality (m)

m = (w_B × 1000) / (M_B × W_A_g)

Unit: mol kg⁻¹ | Temperature independent

W_A = mass of solvent in grams

Mass Percentage

w/w% = (w_B / w_soln) × 100

Unit: % | No temperature dependence

Used in commercial chemical bottles

Henry's Law

p = K_H × χ_gas

K_H unit: bar or atm | Increases with temperature

p = partial pressure; χ = mole fraction of gas

Raoult's Law (Non-volatile solute)

p_s = χ_A × p°_A (p°_A − p_s)/p°_A = χ_B

χ_A = mole fraction of solvent; p°_A = VP of pure solvent

Elevation of Boiling Point

ΔT_b = K_b × m M_B = (K_b × w_B × 1000)/(ΔT_b × w_A)

K_b (water) = 0.52 K kg mol⁻¹

w_A = mass of solvent (g), w_B = mass of solute (g)

Depression of Freezing Point

ΔT_f = K_f × m M_B = (K_f × w_B × 1000)/(ΔT_f × w_A)

K_f (water) = 1.86 K kg mol⁻¹

Most accurate for molar mass determination

Osmotic Pressure

π = CRT = (n/V)RT M_B = (w_B × R × T)/(π × V)

Unit: Pa or atm | R = 8.314 J mol⁻¹K⁻¹ = 0.0821 L atm mol⁻¹K⁻¹

Best for determining molar mass of polymers & proteins

Van't Hoff Factor

i = observed/calculated colligative property i = normal molar mass / abnormal molar mass

i > 1: dissociation | i < 1: association | i = 1: no change

🔗 Relationship Between Molarity & Molality

m = (M × 1000) / (1000ρ − M × M_B)    where ρ = density of solution (g/mL)

💡

Memory Tricks & Mnemonics

Never forget formulas again!

🧠 Remember the 4 Colligative Properties

R E O D

Relative Lowering of Vapour Pressure  •  Elevation of Boiling Point  •  Osmotic Pressure  •  Depression of Freezing Point
"Really Eager Old Dog" 🐕

🧠 Molality vs Molarity — Which is Temperature Independent?

MoLALity = LAst Forever

Molality doesn't change with temperature because it uses mass of solvent (not volume). Mass doesn't change with temperature, but volume does! So Molarity changes (uses volume) and Molality stays (uses mass).

🧠 Van't Hoff Factor — Quick Rule

D = Dissociation → i > 1  |  A = Association → i < 1

Dissociation = Divides into more particles → i > 1 → Decreases molar mass
Association = Adds together into fewer particles → i < 1 → Amplifies molar mass
Remember: "Divide = Dissociate (i>1)" and "Add = Associate (i<1)"

🧠 Ideal Solution Conditions

ZERO ZERO SAME

ΔH_mix = ZERO  |  ΔV_mix = ZERO  |  A–B interactions = SAME as A–A and B–B

🧠 K values for Water — Quick Memory

K_b = 0.52    K_f = 1.86

Trick: K_f (1.86) is almost 4 times K_b (0.52 × 3.58 ≈ 1.86)
"Freezing point is much more sensitive to added solute than boiling point"
Or remember: 52 → Boil, 186 → Freeze (like 52°C summer heat makes you want to freeze at 186!)

🧠 Positive vs Negative Deviation from Raoult's Law

+ve: Ethanol-Water (EW = Escape Wildly) | −ve: Chloroform-Acetone (CA = Clingy Attraction)

+ve deviation: A–B forces WEAKER than A–A/B–B → molecules ESCAPE more easily → higher VP
−ve deviation: A–B forces STRONGER → molecules CLING → lower VP
Positive = Parties (more vapour, more pressure), Negative = No vapour (clingy)

📊

Concept Maps & Mind Maps

Visual summaries for quick learning

🗺️ Chapter Mind Map

╔═══════════════════════════════╗ ║ CHAPTER 1: SOLUTIONS ║ ╚═══════════════╤═══════════════╝ │ ┌──────────────────────┼──────────────────────┐ ▼ ▼ ▼ ┌───────────────┐ ┌─────────────────┐ ┌──────────────────┐ │ TYPES OF │ │ CONCENTRATION │ │ SOLUBILITY │ │ SOLUTIONS │ │ EXPRESSIONS │ │ │ │ │ │ │ │ Henry's Law │ │ Solid/Liq/Gas │ │ • Mole Fraction │ │ p = K_H × χ │ │ in Solid/Liq/ │ │ • Molarity (M) │ │ │ │ Gas (9 types) │ │ • Molality (m) │ │ Effect of T & P │ └───────────────┘ │ • Mass % │ └──────────────────┘ │ • ppm │ └─────────────────┘ │ ┌──────────────────────┼──────────────────────┐ ▼ ▼ ▼ ┌───────────────┐ ┌─────────────────┐ ┌──────────────────┐ │ VAPOUR │ │ IDEAL vs NON- │ │ COLLIGATIVE │ │ PRESSURE & │ │ IDEAL SOLUTIONS │ │ PROPERTIES │ │ RAOULT'S LAW │ │ │ │ │ │ │ │ +ve deviation │ │ 1. RLVP │ │ p_s=χ_A×p°_A │ │ −ve deviation │ │ 2. Elev. of BP │ │ │ │ Azeotropes │ │ 3. Dep. of FP │ └───────────────┘ └─────────────────┘ │ 4. Osmotic Press │ └──────┬───────────┘ │ ┌────────▼────────┐ │ VAN'T HOFF │ │ FACTOR (i) │ │ │ │ Dissociation→i>1│ │ Association→i<1 │ └─────────────────┘

📊 Comparison: Ideal vs Non-Ideal Solutions

Property	Ideal Solution	+ve Deviation	−ve Deviation
Raoult's Law	Obeyed fully ✅	Not obeyed ❌	Not obeyed ❌
Vapour Pressure	P = Pideal	P > Pideal	P < Pideal
ΔHmix	= 0	> 0 (endothermic)	< 0 (exothermic)
ΔVmix	= 0	> 0	< 0
A–B Interactions	= A–A = B–B	< A–A, B–B	> A–A, B–B
Examples	Benzene-Toluene	EtOH-Water	CHCl₃-Acetone

📊 Colligative Properties — Summary Table

Property	Formula	K Value (Water)	Application
RLVP	Δp/p° = χ_B	—	Molar mass determination
Elev. of BP	ΔT_b = K_b × m	K_b = 0.52 K kg/mol	Molar mass (liquids)
Dep. of FP	ΔT_f = K_f × m	K_f = 1.86 K kg/mol	Antifreeze, molar mass
Osmotic Pressure	π = CRT	R = 8.314 J/mol·K	Polymers, desalination

❓

NCERT Important Q&A

Step-by-step solutions

Q1. Calculate the mass percentage of benzene (M=78) and carbon tetrachloride (M=154) if 22 g of benzene is dissolved in 122 g of CCl₄.

Step 1: Total mass of solution = 22 + 122 = 144 g
Step 2: Mass % of benzene = (22/144) × 100 = 15.28%
Step 3: Mass % of CCl₄ = (122/144) × 100 = 84.72%

Answers: Benzene = 15.28%, CCl₄ = 84.72%

Q2. Calculate molarity of 0.25 mol of KCl dissolved in 2 litres of solution.

Formula: M = moles / volume(L)
M = 0.25 / 2 = 0.125 mol L⁻¹

Molarity = 0.125 M

Q3. 1.00 g of non-electrolyte with empirical formula CH₂O is dissolved in 50 g water. The freezing point depression is 0.93°C. Find molar mass. (K_f for water = 1.86 K kg mol⁻¹)

Formula: M_B = (K_f × w_B × 1000) / (ΔT_f × w_A)
M_B = (1.86 × 1 × 1000) / (0.93 × 50)
M_B = 1860 / 46.5 = 40 g/mol
Empirical formula mass of CH₂O = 12+2+16 = 30
n = 40/30 ≈ not a whole number... Actual answer = 180 g/mol with ΔT_f = 0.186°C for glucose (C₆H₁₂O₆)

Molar Mass = 40 g/mol (for the given data)

Q4. Calculate osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming complete dissociation. (M of K₂SO₄ = 174, R = 0.0821 L atm K⁻¹ mol⁻¹)

Moles of K₂SO₄ = 25×10⁻³ / 174 = 1.437 × 10⁻⁴ mol
K₂SO₄ → 2K⁺ + SO₄²⁻, so i = 3
C = 1.437×10⁻⁴ / 2 = 7.187×10⁻⁵ mol/L
π = iCRT = 3 × 7.187×10⁻⁵ × 0.0821 × 298
π = 5.27 × 10⁻³ atm

π = 5.27 × 10⁻³ atm

📚

Previous Year Board Questions

CBSE Board 2018–2024 (with solutions)

1 Mark Questions

CBSE 2023 1 Mark MCQ

Which of the following concentration terms is temperature dependent?
(a) Mole Fraction (b) Molality (c) Molarity (d) Mass Percentage

✅ Answer:

(c) Molarity — because it involves volume of solution, which changes with temperature.

CBSE 20221 Mark

Henry's law constant K_H for CO₂ is 1.67 × 10⁸ Pa at 298K. What is the mole fraction of CO₂ if its partial pressure is 2.5 × 10⁵ Pa?

✅ Answer:

χ = p/K_H = (2.5×10⁵)/(1.67×10⁸) = 1.5 × 10⁻³

2 Mark Questions

CBSE 20242 Marks

What is meant by positive deviation from Raoult's law? Give one example.

✅ Answer:

When the total vapour pressure of a solution is greater than expected from Raoult's law, it is called positive deviation. This occurs when A–B molecular interactions are weaker than A–A and B–B interactions. ΔH_mix > 0 and ΔV_mix > 0.
Example: Ethanol–Water mixture.

CBSE 20232 Marks

State the condition in which Van't Hoff factor i > 1. Give one example.

✅ Answer:

i > 1 when solute undergoes dissociation in solution, producing more particles than formula suggests. Example: NaCl → Na⁺ + Cl⁻ (i = 2 for complete dissociation)

3 Mark Questions

CBSE 20223 Marks

A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if freezing point of pure water is 273.15 K.

✅ Answer:

ΔT_f(sucrose) = 273.15 − 269.15 = 4 K; molality of sucrose: m = (10×1000)/(342×90) = 0.325 mol/kg
K_f = ΔT_f/m = 4/0.325 = 12.3 K kg/mol
Molality of glucose: m = (10×1000)/(180×90) = 0.617 mol/kg
ΔT_f(glucose) = 12.3 × 0.617 = 7.59 K
FP of glucose solution = 273.15 − 7.59 = 265.56 K

5 Mark Questions

CBSE 20195 Marks

(a) State Raoult's Law. How is it similar to Henry's Law? (b) What is meant by 'abnormal molar mass'? (c) Calculate the boiling point of a solution containing 0.61 g of benzoic acid in 50 g of carbon disulphide. (K_b for CS₂ = 2.34 K kg mol⁻¹, M of benzoic acid = 122 g/mol)

✅ Answer:

(a) Raoult's Law: Partial VP of each volatile component = (mole fraction) × (VP of pure component). Henry's Law also states partial pressure ∝ mole fraction of the gas. Both have the same mathematical form; Raoult's law applies to volatile solute, Henry's law to dissolved gases.
(b) When experimentally determined molar mass differs from theoretically expected value due to association or dissociation, it is called abnormal molar mass.
(c) m = (0.61×1000)/(122×50) = 0.1 mol/kg; ΔT_b = 2.34 × 0.1 = 0.234 K
BP of CS₂ = 319.45 K; BP of solution = 319.45 + 0.234 = 319.68 K

Assertion-Reason Questions

CBSE 20241 MarkA-R

Assertion (A): Molality is a better concentration term than molarity for expressing the concentration of solutions.
Reason (R): Molality does not change with change in temperature, but molarity does.

✅ Answer:

(a) Both A and R are correct, and R is the correct explanation of A.
Molality uses mass of solvent (which doesn't change with T), so it's temperature independent. Molarity uses volume (which expands/contracts with T).

Case-Based Question

CBSE 20234 MarksCase Based

Read the following passage and answer:
Osmosis plays a crucial role in biological systems. Red blood cells placed in pure water swell and burst (haemolysis) due to endosmosis. When placed in concentrated NaCl solution, they shrink (crenation) due to exosmosis. This is why hospitals use 0.9% NaCl (normal saline) for IV drips — it is isotonic with blood plasma (osmotic pressure ≈ 7.4 atm at 37°C).

(i) What is meant by isotonic solutions?
(ii) Calculate the concentration of glucose (M=180) solution that would be isotonic with 0.9% NaCl solution. (M of NaCl = 58.5, assume complete dissociation)
(iii) What happens when a plant cell is placed in a hypertonic solution?
(iv) Name the process used in industries to purify sea water.

✅ Answer:

(i) Isotonic solutions have equal osmotic pressures; no net osmosis occurs between them.
(ii) Moles of NaCl in 0.9g/100mL = 0.9/58.5 × 10 = 0.154 mol/L; i = 2; π_NaCl = 2 × 0.154 × RT; For isotonic glucose: C = 0.308 mol/L; mass = 0.308 × 180 = 55.4 g/L ≈ 5.5%
(iii) The cell loses water due to exosmosis and undergoes plasmolysis.
(iv) Reverse Osmosis (RO)

🎯

Most Important Exam Questions

30 frequently asked questions — must practice!

#	Question	Marks	Frequency
1	Define Raoult's Law. State its two limitations.	2	⭐⭐⭐⭐⭐
2	Distinguish between ideal and non-ideal solutions.	3	⭐⭐⭐⭐⭐
3	What are colligative properties? List all four with formulas.	3	⭐⭐⭐⭐⭐
4	Derive expression for relative lowering of vapour pressure.	3	⭐⭐⭐⭐
5	Define Van't Hoff factor. Give its value for NaCl, K₂SO₄, AlCl₃.	2	⭐⭐⭐⭐⭐
6	Why is molality preferred over molarity for colligative properties?	2	⭐⭐⭐⭐
7	State Henry's Law and give three applications.	3	⭐⭐⭐⭐
8	Calculate molar mass using depression of freezing point.	3–5	⭐⭐⭐⭐⭐
9	A solution has elevation of boiling point 0.52°C. Find molality.	2	⭐⭐⭐⭐
10	What is osmotic pressure? How is it used to find molar mass?	3	⭐⭐⭐⭐
11	Explain why is the freezing point of sea water lower than pure water.	2	⭐⭐⭐
12	What happens when we add ethylene glycol to car radiator water?	2	⭐⭐⭐
13	Benzoic acid in benzene shows i < 1. Explain.	2	⭐⭐⭐⭐
14	Explain the concept of reverse osmosis with application.	3	⭐⭐⭐⭐
15	What is a semi-permeable membrane? Give two examples.	1–2	⭐⭐⭐
16	Define K_b and K_f. State their SI units.	2	⭐⭐⭐⭐
17	Two liquids A and B form ideal solution. Calculate VP at 300K.	3	⭐⭐⭐⭐
18	Define mole fraction. Why is sum of all mole fractions always = 1?	2	⭐⭐⭐
19	What is azeotrope? Distinguish minimum and maximum boiling azeotrope.	3	⭐⭐⭐⭐
20	Why do gases always show positive deviation from Raoult's law?	2	⭐⭐⭐
21	Numerical: Osmotic pressure of haemoglobin solution at 25°C.	3	⭐⭐⭐⭐
22	Numerical: Molarity and Molality of 18% by mass H₂SO₄ (d=1.2 g/mL).	3	⭐⭐⭐⭐⭐
23	Define i for KCl. If degree of dissociation = 0.9, find i.	2	⭐⭐⭐⭐
24	Calculate mole fraction of ethanol in 20% (w/w) aqueous solution.	2	⭐⭐⭐⭐
25	What is the effect of temperature on Henry's law constant? Explain.	2	⭐⭐⭐
26	Derive the relationship between observed and normal molar mass using i.	3	⭐⭐⭐
27	Numerical: 0.5 g NaCl in 100 g water; find ΔT_f. (i=2, K_f=1.86)	2	⭐⭐⭐⭐⭐
28	What is isotonic, hypertonic, and hypotonic solution?	3	⭐⭐⭐⭐
29	Write three differences between molarity and molality.	3	⭐⭐⭐⭐
30	Explain why ΔT_f > ΔT_b for the same molal solution of a solute in water.	2	⭐⭐⭐

🧮

Solved Numericals

Step-by-step with shortcut methods

N1. Calculate Molarity and Molality of a 30% (w/w) H₂O₂ solution. Density = 1.11 g/mL. (M of H₂O₂ = 34)

Assume 100 g of solution: 30 g H₂O₂, 70 g H₂O
Moles of H₂O₂ = 30/34 = 0.882 mol
Volume of solution = 100/1.11 = 90.09 mL = 0.09009 L
Molarity M = 0.882/0.09009 = 9.79 mol/L ≈ 9.8 M
Molality m = (0.882 × 1000)/70 = 12.6 mol/kg

M = 9.8 mol/L | m = 12.6 mol/kg

N2. The vapour pressure of pure water at 25°C is 3.167 kPa. Find the vapour pressure of the solution when 10 g of glucose (M=180) is dissolved in 90 g of water.

Moles of glucose (n_B) = 10/180 = 0.0556 mol
Moles of water (n_A) = 90/18 = 5 mol
χ_B = 0.0556/(0.0556+5) = 0.0556/5.0556 = 0.011
χ_A = 1 − 0.011 = 0.989
p_s = χ_A × p°_A = 0.989 × 3.167 = 3.132 kPa

VP of solution = 3.132 kPa

N3. Find the boiling point of a solution containing 0.456 g of camphor (M=152) dissolved in 31.4 g of acetone. K_b for acetone = 1.72 K kg mol⁻¹. BP of acetone = 329.45 K.

m = (0.456 × 1000)/(152 × 31.4) = 456/4772.8 = 0.0956 mol/kg
ΔT_b = K_b × m = 1.72 × 0.0956 = 0.164 K
BP of solution = 329.45 + 0.164 = 329.61 K

BP = 329.61 K

N4. An aqueous solution of urea has a boiling point of 100.18°C. If K_b for water = 0.512 K kg/mol and K_f = 1.86 K kg/mol, calculate the freezing point of the same solution.

ΔT_b = 100.18 − 100 = 0.18°C
m = ΔT_b/K_b = 0.18/0.512 = 0.3516 mol/kg
ΔT_f = K_f × m = 1.86 × 0.3516 = 0.654°C
FP of solution = 0 − 0.654 = −0.654°C = 272.496 K

Freezing point = −0.654°C

N5. 200 cm³ of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure at 300K is 2.57 × 10⁻³ bar. Find molar mass. (R = 83.1 cm³ bar K⁻¹ mol⁻¹)

π = CRT → C = π/RT = (2.57×10⁻³)/(83.1×300) = 1.031×10⁻⁷ mol/cm³
Moles = 1.031×10⁻⁷ × 200 = 2.062×10⁻⁵ mol
M = 1.26/(2.062×10⁻⁵) = 61,106 g/mol ≈ 61.1 kg/mol

Molar Mass ≈ 61,086 g/mol (suitable for a protein)

N6. (Abnormal Molar Mass) 0.6 g of acetic acid dissolved in 1 kg benzene shows a freezing point depression of 0.0108°C. K_f = 5.12 K kg mol⁻¹. Find observed molar mass and degree of association.

M_observed = (K_f × w_B × 1000)/(ΔT_f × w_A) = (5.12 × 0.6 × 1000)/(0.0108 × 1000) = 3072/10.8 = 284.4 g/mol
Normal MM of CH₃COOH = 60 g/mol
i = M_normal/M_observed = 60/284.4 = 0.211
Acetic acid dimerizes: 2CH₃COOH ⇌ (CH₃COOH)₂
If α = degree of association: i = 1 − α/2
0.211 = 1 − α/2 → α/2 = 0.789 → α = 0.943 ≈ 94.3%

M_observed = 284.4 g/mol | Degree of association ≈ 94.3%

🧠

HOTS Questions

Higher Order Thinking Skills — challenge yourself!

H1. A student dissolved NaCl in water and measured its boiling point. She then dissolved the same mass of glucose in water and measured its boiling point. Which solution boils at a higher temperature and why?

💡 Hint: ΔT_b = i × K_b × m. NaCl dissociates (i=2), glucose doesn't (i=1). For same mass: molality of glucose (M=180) < NaCl (M=58.5). Calculate both i×m products. NaCl gives higher ΔT_b because i × (mass/M) is larger for NaCl despite lower molality.

H2. Why is it not advisable to use animal cells (like RBCs) in osmosis experiments if the concentration of external solution is very high?

💡 Hint: In hypertonic solution (high concentration outside), water leaves cells by exosmosis (crenation). This shrivels and destroys cells. However, too dilute a solution causes haemolysis (cell bursts). Only isotonic solutions maintain cell integrity — this is why 0.9% NaCl is used in medicine.

H3. The Henry's law constant K_H for a gas increases with temperature. What does this mean for the solubility of that gas?

💡 Hint: From p = K_H × χ → χ = p/K_H. As K_H increases (with T), χ decreases for same pressure. So solubility decreases with temperature. This explains why cold drinks go flat when warm — CO₂ escapes as temperature rises, because K_H for CO₂ increases with T.

H4. Explain why the osmotic method is preferred over boiling point elevation for determining molar mass of polymers and proteins.

💡 Hint: High molar mass → very low molality → very small ΔT_b (unmeasurable). But osmotic pressure is much larger (π ≈ 8.3 kPa for even dilute polymer solution at 300 K). Example: 1 g/L protein (M~60,000) gives π ≈ 0.04 atm but ΔT_b ≈ 0.00009°C. Osmometry is ~1000× more sensitive.

H5. If two solutions A and B have the same osmotic pressure at the same temperature but contain different solutes, what can you conclude about them?

💡 Hint: They are isotonic. Since π = iCRT, equal π means equal effective concentrations (iC). However, actual molar concentrations may differ if van't Hoff factors differ. A non-electrolyte at 1 M has same π as 0.5 M NaCl (i=2).

🚫

Common Mistakes to Avoid

Don't lose easy marks — read carefully!

❌

Confusing Molarity and Molality formulas Molarity uses volume of solution (L); Molality uses mass of solvent (kg). Many students write wrong denominators and lose all marks.

❌

Forgetting Van't Hoff Factor in numerical problems For electrolytes, always multiply by i. NaCl: i=2, K₂SO₄: i=3, AlCl₃: i=4. Forgetting i is the #1 error in colligative property numericals.

❌

Wrong formula for molar mass from K_b or K_f Many use M_B = K_f × w_B / (ΔT_f × W_A). Watch out: w_A must be in grams and the formula has ×1000 to convert to kg. Always write the full formula clearly.

❌

Stating that ideal solutions have no intermolecular forces Wrong! Ideal solutions have A–A, B–B, and A–B interactions that are equal in magnitude, not zero. The key condition is ΔH_mix = 0 and ΔV_mix = 0.

❌

Confusing osmosis direction Osmosis: solvent moves from LOW concentration (high water potential) to HIGH concentration (low water potential). Not solute — always refer to solvent movement!

❌

Unit error in osmotic pressure formula π = CRT. Use consistent units: if C in mol/L and R = 0.0821 L·atm/mol·K, answer in atm. If R = 8.314 J/mol·K, C must be in mol/m³ → answer in Pa.

❌

Saying Raoult's Law and Henry's Law are the same They have same mathematical form but differ in scope. Raoult's law: applies to solvents/volatile components. Henry's law: applies to dissolved gases. K_H ≠ p°.

❌

Forgetting that ΔT_f = T_f(pure) − T_f(solution) ΔT_f is always positive (a decrease). Do NOT write T_f(solution) = T_f(pure) + ΔT_f. The freezing point goes DOWN: T_f(solution) = T_f(pure) − ΔT_f.

📋

MCQ Practice — 35 Questions

Click "Show Answer" to reveal the correct option

1. Which of the following is a colligative property?

(a) Viscosity

(b) Surface tension

(c) Osmotic pressure

(d) Optical rotation

2. The unit of Van't Hoff factor (i) is:

(a) mol L⁻¹

(b) mol kg⁻¹

(c) Dimensionless

(d) atm

3. Molality of a solution is independent of temperature because:

(a) It involves mass of solution

(b) It involves mass of solvent

(c) It involves volume of solution

(d) It involves density of solution

4. For an electrolyte NaCl completely dissociated, the Van't Hoff factor is:

(a) 0.5

(b) 1

(c) 2

(d) 3

5. Ebullioscopic constant (K_b) depends on:

(a) Nature of solute

(b) Nature of solvent

(c) Concentration of solution

(d) Temperature of solution

6. Which solution would have the maximum boiling point?

(a) 1 M glucose

(b) 1 M NaCl

(c) 1 M BaCl₂

(d) 1 M AlCl₃

7. Henry's law is applicable to:

(a) Solid in liquid solutions

(b) Gas in gas solutions

(c) Gas in liquid solutions

(d) Liquid in liquid solutions

8. What happens to vapour pressure when a non-volatile solute is dissolved?

(a) Increases

(b) Decreases

(c) Remains same

(d) First increases then decreases

9. A solution of benzoic acid in benzene shows i = 0.5. This is due to:

(a) Ionisation

(b) Hydrolysis

(c) Dimerization

(d) Decomposition

10. Reverse osmosis is used in:

(a) Dialysis in kidneys

(b) Desalination of sea water

(c) Blood transfusion

(d) Drug delivery

11. For an ideal solution, ΔH_mix is:

(a) Positive

(b) Negative

(c) Zero

(d) Infinite

12. The value of K_f for water is:

(a) 0.512 K kg mol⁻¹

(b) 1.86 K kg mol⁻¹

(c) 2.53 K kg mol⁻¹

(d) 3.9 K kg mol⁻¹

13. Which of these concentration terms has unit mol kg⁻¹?

(a) Molarity

(b) Normality

(c) Mole Fraction

(d) Molality

14. Osmosis is the flow of:

(a) Solute from high to low concentration

(b) Solvent from high to low solvent concentration

(c) Solvent from low to high solute concentration

(d) Both solute and solvent

15. The sum of mole fractions of all components in a solution is always:

(a) 0

(b) 0.5

(c) 1

(d) Depends on number of components

16. The process used for purification of sea water by applying pressure greater than osmotic pressure is:

(a) Dialysis

(b) Osmosis

(c) Reverse Osmosis

(d) Distillation

17. An aqueous solution that freezes at −0.186°C has a molality of: (K_f = 1.86 K kg/mol)

(a) 0.001 mol/kg

(b) 0.1 mol/kg

(c) 0.01 mol/kg

(d) 1 mol/kg

18. Azeotropes are formed by:

(a) Ideal solutions only

(b) Non-ideal solutions

(c) Dilute solutions

(d) Any solution

19. Ethanol–water mixture shows:

(a) Negative deviation from Raoult's law

(b) Positive deviation from Raoult's law

(c) Ideal behaviour

(d) No deviation

20. Which of the following is the best solvent for determining molar mass of an unknown compound by freezing point method?

(a) Water (K_f = 1.86)

(b) Benzene (K_f = 5.12)

(c) Camphor (K_f = 40)

(d) Ethanol (K_f = 1.99)

📝 MCQs 21–35 — Answer Key

#	Question	Answer
21	K₂SO₄ completely dissociates. Its Van't Hoff factor is:	(d) 3 — 2K⁺ + SO₄²⁻
22	Which has highest osmotic pressure at same concentration? (NaCl, KCl, glucose, sucrose)	NaCl or KCl (i=2); glucose/sucrose i=1
23	ppm is used to express concentration of:	(b) Very dilute solutions
24	The elevation of boiling point is a colligative property because it depends on:	(c) Number of solute particles
25	Haemolysis of RBCs occurs when placed in:	(a) Hypotonic solution (less concentrated)
26	If degree of dissociation of MX₂ is α, its i value is:	1 + 2α
27	For an ideal binary solution (A+B), if χ_A = 0.4, p°_A = 80 mm, p°_B = 60 mm, total VP =	68 mm Hg (0.4×80 + 0.6×60)
28	π = CRT equation is analogous to:	PV = nRT (ideal gas equation)
29	Mass percentage of oxygen in water (M=18):	88.88%
30	Relative lowering of VP = χ_B means it is independent of:	Nature of solute (only depends on mole fraction)
31	Minimum boiling azeotrope shows what type of deviation?	Positive deviation
32	At higher altitudes, solubility of CO₂ in water:	Decreases (lower partial pressure)
33	Cryoscopic constant depends on:	Nature of solvent only
34	For 0.1 m NaCl (i=2) and 0.1 m glucose: which has higher ΔT_b?	NaCl (i=2 vs i=1)
35	Semi-permeable membrane allows passage of:	Solvent molecules only

🚀

Quick Revision Notes

Last-minute preparation — 1 page summary

📏 Concentration

• χ = n_A/(n_A+n_B)
• M = mol/L (T-dependent)
• m = mol/kg (T-independent ✅)
• Mass% = (w_B/w_soln)×100
• ppm = mg/kg

💨 Vapour Pressure

• p_s = χ_A × p°_A
• RLVP = χ_B (Raoult)
• Henry: p = K_H × χ
• +ve dev → p > ideal
• −ve dev → p < ideal

🌡️ Colligative

• ΔT_b = K_b × m (K_b=0.52)
• ΔT_f = K_f × m (K_f=1.86)
• π = CRT
• With i: multiply each by i

🔬 Van't Hoff

• i = obs/calc property
• Dissociation: i > 1
• Association: i < 1
• NaCl: i=2, K₂SO₄: i=3
• Benzoic acid/Benzene: i=0.5

⭐ Key Values

• K_b (H₂O) = 0.52
• K_f (H₂O) = 1.86
• K_f (benzene) = 5.12
• K_f (camphor) = 40
• R = 8.314 J/mol·K

🧪 Examples

• Ideal: benzene-toluene
• +ve: EtOH-H₂O
• −ve: CHCl₃-acetone
• Min BP azeotrope: +ve
• Max BP azeotrope: −ve

🏆

Board Exam Statistics

Chapter 1 (Solutions) typically carries 8–12 marks in CBSE Board exams. Questions appear as: 1 MCQ (1M), 1 VSA (2M), 1 SA (3M), 1 numerical (3M), and sometimes 1 long answer (5M). Practicing all formula-based numericals guarantees full marks in this chapter!

🎯

Board Exam Preparation Tips

Score maximum marks from this chapter

1

Master the formulas first. Write all formulas on a single page and revise daily. Pay special attention to which constant goes with which property (K_b for ΔT_b, K_f for ΔT_f).

2

Always check Van't Hoff factor. In any numerical involving electrolytes, immediately check if i should be applied. Forgetting i = losing 1–2 marks per numerical.

3

Show complete steps in numericals. Even if your final answer is wrong, step marking ensures partial credit. Write formula → substitution → calculation → answer with units.

4

Learn at least 2 examples for each type. For theory questions asking "give example of ideal solution", examiners want specific pairs like benzene-toluene or n-hexane–n-heptane, not vague answers.

5

Draw labelled diagrams where asked. For osmosis, drawing a U-tube diagram with labelled SPM, solution levels, and arrows showing flow earns extra marks.

6

Revise definitions precisely. Definitions must be verbatim or very close. Use keyword triggers: Raoult's Law (partial pressure), Henry's Law (partial pressure of gas), osmotic pressure (minimum excess pressure).

7

Practice numerical from different angles. Boards can ask: given ΔT_f find M_B, or given M find ΔT_f. Practice both ways so formula manipulation is automatic.

8

Attempt MCQs last in the section. MCQs in board exams are worth 1 mark each. Don't spend more than 1 minute per MCQ. Use elimination method and move on.

🌟

Chapter Summary

Everything you need to know — at a glance

📌 Solutions Basics

• Homogeneous mixture of solute + solvent
• 9 types based on physical state
• Binary solution = 2 components
• Saturated, unsaturated, supersaturated

📌 Concentration Terms

• Mole fraction, mass%, molarity, molality
• Only molality is T-independent
• Sum of all mole fractions = 1
• ppm used for trace amounts

📌 Raoult's Law

• p_i = χ_i × p°_i for each component
• Ideal: ΔH=0, ΔV=0
• +ve deviation: weaker A–B forces
• −ve deviation: stronger A–B forces

📌 Colligative Properties

• 4 types: RLVP, ΔT_b, ΔT_f, π
• Depend only on NUMBER of particles
• Used to find molar mass
• Modified by Van't Hoff factor i

📌 Osmosis

• Solvent through SPM: dilute → concentrated
• Isotonic: equal osmotic pressure
• Reverse osmosis: purifies sea water
• π = CRT (van't Hoff equation)

📌 Van't Hoff Factor

• i = observed/normal molar mass ratio
• Dissociation: i > 1
• Association: i < 1
• Normal (no change): i = 1

🙏 You've Got Everything You Need!

Dear student, you've just covered one of the most important chapters in Class 12 Chemistry. Remember — Chemistry is not about memorization, it is about understanding. When you truly understand WHY a non-volatile solute raises the boiling point and lowers the freezing point, the formula becomes obvious.

Here is your daily action plan for the next 2 weeks:

📅 Day 1–2: Read concepts & definitions  |  📅 Day 3–4: Memorize formulas  |  📅 Day 5–7: Solve all numericals  |  📅 Day 8–10: Practice PYQs  |  📅 Day 11–14: Revise & solve MCQs

"Success in board exams is not about studying more — it is about studying smart. You have this guide, now go use it!"

Best of luck from Jnaanangkur – The Learning Hub! 🌱

📚 Jnaanangkur – The Learning Hub | Class 12 Chemistry | Chapter 1: Solutions

NCERT Aligned | CBSE & State Board | Last Updated: 2024–25 Academic Year

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