CBSE Class 11 Chemistry Mole Concept Notes, Questions & Answers (NEET + Board Exam)

 - Dr Sanjay Kumar Pawar 

CBSE Class 11 Chemistry Mole Concept Notes PDF | MCQs, Questions & Answers

Class 11 Chemistry Mole Concept Complete Guide: Formulas, Examples, MCQs and Practice Questions for CBSE and NEET Students

 Internal Links

Chemistry Chapter Links

1. Basic Concepts of Chemistry

Basic Concepts of Chemistry Class 11 Notes

URL:

/basic-concepts-of-chemistry-class-11

---

2. Atomic Structure Notes

Class 11 Atomic Structure Complete Notes

URL:

/atomic-structure-class-11-chemistry

---

3. Periodic Table Notes

Periodic Classification of Elements Notes

URL:

/periodic-table-class-11

---

4. Chemical Bonding Notes

Chemical Bonding Class 11 Chemistry

URL:

/chemical-bonding-class-11

---

5. Stoichiometry Questions

Stoichiometry Numericals and Practice Questions

URL:

/stoichiometry-class-11-chemistry

---

6. NEET Chemistry Preparation

NEET Chemistry Notes and Practice Questions

URL:

/neet-chemistry-notes

---

7. CBSE Class 11 Chemistry Question Bank

CBSE Class 11 Chemistry Question Bank

URL:

/cbse-class-11-chemistry-question-bank

---

CBSE Class 11 Chemistry

Chapter 1: Some Basic Concepts of Chemistry

Mole Concept Question Bank

Contents of Part 1A

• Chapter Introduction
• Learning Outcomes
• MCQs (1–10)

Chapter Introduction

The Mole Concept is one of the most important topics in Chemistry. It helps us calculate the amount of substances involved in chemical reactions. The concept is based on Avogadro's Number (6.022 × 10²³), which represents the number of particles in one mole of any substance.

Learning Outcomes

• Understand the meaning of mole.
• Learn Avogadro's number.
• Calculate molar mass.
• Convert mass into moles.
• Calculate number of particles.
• Solve numerical problems.

Section A : Multiple Choice Questions (MCQs)

Q1. One mole contains

A. 6.022 × 10²² particles
B. 3.011 × 10²³ particles
C. 6.022 × 10²³ particles
D. 12 × 10²³ particles

Answer: C

Q2. SI unit of amount of substance is

A. Gram
B. Mole
C. Kilogram
D. Liter

Answer: B

Q3. Avogadro Number is

A. 6.022 × 10²²
B. 6.022 × 10²³
C. 22.4
D. 96500

Answer: B

Q4. Molar mass of water is

A. 16 g/mol
B. 17 g/mol
C. 18 g/mol
D. 20 g/mol

Answer: C

Q5. One mole of oxygen molecules contains

A. 6.022 × 10²³ molecules
B. 3.011 × 10²³ molecules
C. 12.044 × 10²³ molecules
D. 22.4 molecules

Answer: A

Q6. Molar mass of CO₂ is

A. 28 g/mol
B. 32 g/mol
C. 44 g/mol
D. 46 g/mol

Answer: C

Q7. One mole of gas occupies at STP

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 18 L

Answer: B

Q8. One Faraday is equal to

A. 9650 C
B. 96500 C
C. 965000 C
D. 6022 C

Answer: B

Q9. Number of moles in 18 g of water is

A. 0.5
B. 1
C. 2
D. 3

Answer: B

Q10. Which of the following contains the same number of molecules?

A. 1 mole H₂
B. 1 mole O₂
C. 1 mole CO₂
D. All of these

Answer: D

End of Part 1A (Continue with Part 1B)

Section A: Multiple Choice Questions (MCQs 11–25)

Q11. Molar mass of NaCl is

A. 23 g/mol
B. 35.5 g/mol
C. 58.5 g/mol
D. 46 g/mol

Answer: C

Q12. One mole of carbon atoms has a mass of

A. 1 g
B. 12 g
C. 16 g
D. 24 g

Answer: B

Q13. Number of atoms present in one mole of carbon is

A. 6.022 × 10²²
B. 3.011 × 10²³
C. 6.022 × 10²³
D. 12 × 10²³

Answer: C

Q14. The molar mass of oxygen gas (O₂) is

A. 16 g/mol
B. 18 g/mol
C. 32 g/mol
D. 44 g/mol

Answer: C

Q15. Number of moles in 44 g of CO₂ is

A. 0.5
B. 1
C. 2
D. 4

Answer: B

Q16. Formula for calculating moles is

A. Mass × Molar Mass
B. Mass ÷ Molar Mass
C. Molar Mass ÷ Mass
D. Mass + Molar Mass

Answer: B

Q17. Which quantity is represented by the symbol NA?

A. Atomic Number
B. Avogadro Constant
C. Atomic Mass
D. Number of Electrons

Answer: B

Q18. Number of moles present in 23 g of sodium is

A. 0.5
B. 1
C. 2
D. 3

Answer: B

Q19. Which one is an SI unit?

A. Gram
B. Mole
C. Litre
D. Millilitre

Answer: B

Q20. One mole of electrons carries

A. 965 C
B. 9650 C
C. 96500 C
D. 965000 C

Answer: C

Q21. The molar mass of NH₃ is

A. 14 g/mol
B. 15 g/mol
C. 17 g/mol
D. 18 g/mol

Answer: C

Q22. One mole of hydrogen molecules contains

A. 6.022 × 10²³ molecules
B. 6.022 × 10²² molecules
C. 12.044 × 10²³ molecules
D. 3.011 × 10²³ molecules

Answer: A

Q23. The atomic mass of oxygen is

A. 8 u
B. 12 u
C. 14 u
D. 16 u

Answer: D

Q24. Which of the following has the highest molar mass?

A. H₂O
B. NH₃
C. CO₂
D. CH₄

Answer: C

Q25. One mole of NaCl contains

A. 6.022 × 10²³ formula units
B. 22.4 molecules
C. 96500 ions
D. 58.5 atoms

Answer: A

---

Section B: Fill in the Blanks

1. One mole contains __________ particles.
   Answer: 6.022 × 10²³


2. The SI unit of amount of substance is __________.
   Answer: Mole


3. The mass of one mole of a substance is called __________.
   Answer: Molar Mass


4. One mole of gas occupies __________ litres at STP.
   Answer: 22.4


5. Avogadro constant is represented by __________.
   Answer: NA


6. One Faraday is equal to __________ coulomb.
   Answer: 96500


7. Moles = Mass ÷ __________.
   Answer: Molar Mass


8. Mass = Moles × __________.
   Answer: Molar Mass


9. The molar mass of CO₂ is __________ g/mol.
   Answer: 44


10. The molar mass of H₂O is __________ g/mol.
    Answer: 18

End of Part 1B
Continue with Part 1C (True/False + Very Short Answer Questions + Closing HTML)

Section C : True / False

1. One mole contains 6.022 × 10²³ particles.

Answer: True

2. Mole is the SI unit of amount of substance.

Answer: True

3. One mole of gas occupies 44.8 L at STP.

Answer: False (22.4 L)

4. Molar mass is expressed in g/mol.

Answer: True

5. One Faraday equals 96500 C.

Answer: True

6. Water has a molar mass of 20 g/mol.

Answer: False (18 g/mol)

7. One mole of carbon contains Avogadro number of atoms.

Answer: True

8. The molar mass of CO₂ is 44 g/mol.

Answer: True

9. The atomic mass of oxygen is 32 u.

Answer: False (16 u)

10. Mass = Moles × Molar Mass.

Answer: True

---

Section D : Very Short Answer Questions (1 Mark)

Q1. Define mole.

A mole is the amount of substance containing 6.022 × 10²³ particles.

Q2. What is Avogadro's Number?

6.022 × 10²³ particles per mole.

Q3. What is molar mass?

Mass of one mole of a substance expressed in g/mol.

Q4. Write the SI unit of amount of substance.

Mole

Q5. State the molar volume of a gas at STP.

22.4 L

Q6. Write the formula for calculating moles.

Moles = Mass ÷ Molar Mass

Q7. What is one Faraday?

The charge carried by one mole of electrons (96500 C).

Q8. What is the molar mass of oxygen gas?

32 g/mol

Q9. What is the molar mass of sodium chloride?

58.5 g/mol

Q10. How many molecules are present in one mole of water?

6.022 × 10²³ molecules.

---

Quick Formula Sheet

Formula	Expression
Moles	Mass ÷ Molar Mass
Mass	Moles × Molar Mass
Particles	Moles × 6.022 × 10²³
Moles	Particles ÷ 6.022 × 10²³
Gas Volume at STP	Moles × 22.4 L
Charge	Moles × 96500 C

Part 1 Completed Successfully

• ✓ Chapter Introduction
• ✓ Learning Outcomes
• ✓ 25 MCQs with Answers
• ✓ Fill in the Blanks
• ✓ True / False Questions
• ✓ Very Short Answer Questions
• ✓ Formula Sheet

CBSE Class 11 Chemistry

Chapter 1 – Some Basic Concepts of Chemistry (Mole Concept)

End of Part 1

Continue with Part 2 (Short Answer, Long Answer, Numericals, Assertion & Reason, Statement-Based Questions).

CBSE Class 11 Chemistry

Chapter 1: Some Basic Concepts of Chemistry

Mole Concept

Part 2A-1a

Short Answer Questions (2–3 Marks)

Short Answer Questions (Questions 1–10)

Question 1

Define mole. Why is it called the SI unit of amount of substance?

Answer:

A mole is the amount of substance that contains 6.022 × 10²³ particles (atoms, molecules, ions, or electrons). It is the SI unit used to measure the amount of a chemical substance.

Question 2

What is Avogadro's number? Mention its importance.

Answer:

Avogadro's number is 6.022 × 10²³ particles per mole. It is important because it relates the microscopic world (atoms and molecules) with the macroscopic world (mass measured in grams).

Question 3

Differentiate between atomic mass and molar mass.

Answer:

Atomic Mass	Molar Mass
Mass of one atom	Mass of one mole of substance
Unit: u (amu)	Unit: g/mol
Very small quantity	Can be measured experimentally

Question 4

Calculate the number of moles present in 18 g of water.

Answer:

Mass = 18 g
Molar Mass of Water = 18 g/mol

Moles = Mass ÷ Molar Mass
= 18 ÷ 18
= 1 mole

Question 5

How many molecules are present in one mole of carbon dioxide?

Answer:

One mole of CO₂ contains 6.022 × 10²³ molecules.

Question 6

State any three applications of the mole concept.

Answer:

1. Calculation of masses of substances.
2. Calculation of number of atoms and molecules.
3. Chemical equation and stoichiometric calculations.

Question 7

Write the formula for calculating moles from mass and explain each term.

Answer:

Formula:

Moles = Mass ÷ Molar Mass

• Mass = Amount of substance in grams
• Molar Mass = Mass of one mole (g/mol)
• Moles = Amount of substance

Question 8

What is molar volume? Write its value at STP.

Answer:

Molar volume is the volume occupied by one mole of any gas. At STP, 1 mole of gas occupies 22.4 L.

Question 9

Calculate the molar mass of carbon dioxide (CO₂).

Answer:

Atomic Mass of Carbon = 12
Atomic Mass of Oxygen = 16

CO₂ = 12 + (16 × 2)

= 44 g/mol

Question 10

Calculate the volume occupied by 2 moles of oxygen gas at STP.

Answer:

Volume of one mole gas = 22.4 L

Volume = Number of moles × 22.4

= 2 × 22.4

= 44.8 L

End of Part 2A-1a

Next: Part 2A-1b (Questions 11–20)

Short Answer Questions (2–3 Marks)

Question 11

What is molar volume? State its value at STP.

Answer:

• Molar volume is the volume occupied by one mole of any gas at STP.
• Its value is 22.4 L.

---

Question 12

Calculate the number of molecules present in 18 g of water.

Answer:

Molar mass of water = 18 g/mol

Number of moles = 18 ÷ 18 = 1 mol

Number of molecules = 1 × 6.022 × 10²³

= 6.022 × 10²³ molecules

---

Question 13

What is empirical formula?

Answer:

The empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound.

Example:

Glucose (C₆H₁₂O₆) Empirical Formula = CH₂O

---

Question 14

What is molecular formula?

Answer:

The molecular formula shows the actual number of atoms of each element present in one molecule of a compound.

Example:

Water = H₂O Glucose = C₆H₁₂O₆

---

Question 15

Differentiate between empirical formula and molecular formula.

Empirical Formula	Molecular Formula
Simplest ratio of atoms	Actual number of atoms
May not represent one molecule	Represents one molecule
Example: CH₂O	Example: C₆H₁₂O₆

---

Question 16

State the law of conservation of mass.

Answer:

Mass can neither be created nor destroyed during a chemical reaction. The total mass of reactants is equal to the total mass of products.

---

Question 17

State the law of constant proportions.

Answer:

A pure chemical compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source.

---

Question 18

Calculate the molar mass of sulphuric acid (H₂SO₄).

Answer:

H = 1 × 2 = 2 S = 32 × 1 = 32 O = 16 × 4 = 64

Total = 2 + 32 + 64

= 98 g/mol

---

Question 19

Calculate the number of moles present in 49 g of H₂SO₄.

Answer:

Molar mass = 98 g/mol

Moles = 49 ÷ 98

= 0.5 mol

---

Question 20

Write any four applications of the mole concept.

Answer:

1. Calculation of mass of substances.
2. Calculation of number of particles.
3. Calculation of gas volume.
4. Balancing and solving chemical equations.

---

Quick Revision

• 1 Mole = 6.022 × 10²³ particles
• Moles = Mass ÷ Molar Mass
• Mass = Moles × Molar Mass
• Particles = Moles × 6.022 × 10²³
• 1 Mole Gas = 22.4 L at STP
• 1 Faraday = 96500 C

---

CBSE Class 11 Chemistry
Part 2A-1 Completed (Questions 11–20)
Chapter: Some Basic Concepts of Chemistry (Mole Concept)

CBSE Class 11 Chemistry

Mole Concept - Part 2A-2.1

Short Answer Questions (2–3 Marks)

Questions 21–30

Q21. Explain the relationship between mole and Avogadro's number.

One mole of any substance contains exactly 6.022 × 10²³ particles. This number is called Avogadro's number (NA). It helps convert between moles and the number of particles.

Q22. Calculate the number of moles present in 90 g of water.

Molar mass of H₂O = 18 g/mol

Moles = Mass ÷ Molar Mass
= 90 ÷ 18
= 5 mol

Q23. Calculate the molar mass of ammonia (NH₃).

Atomic mass of Nitrogen = 14
Atomic mass of Hydrogen = 1

Molar mass
= 14 + (3 × 1)
= 17 g/mol

Q24. Why is the mole concept important in chemistry?

The mole concept helps in:

• Counting atoms and molecules.
• Calculating masses of substances.
• Performing chemical calculations.
• Balancing chemical equations.

Q25. Find the number of molecules in one mole of carbon dioxide.

Number of molecules
= 1 × 6.022 × 10²³

= 6.022 × 10²³ molecules

Q26. State the formula for calculating moles from mass.

Formula:

Moles = Mass ÷ Molar Mass

Unit of molar mass = g/mol

Q27. What is the volume occupied by 3 moles of oxygen gas at STP?

Volume = Number of moles × 22.4

= 3 × 22.4
= 67.2 L

Q28. Calculate the number of atoms present in one mole of helium.

One mole contains Avogadro's number of atoms.

Number of atoms
= 6.022 × 10²³

Q29. Differentiate between atoms and molecules.

Atom	Molecule
Smallest particle of an element.	Two or more atoms chemically combined.
Cannot be divided chemically.	Can contain same or different atoms.
Example: He	Example: H₂O

Q30. Calculate the mass of 2.5 moles of sodium chloride (NaCl).

Molar mass of NaCl
= 23 + 35.5
= 58.5 g/mol

Mass
= Moles × Molar Mass
= 2.5 × 58.5
= 146.25 g

End of Part 2A-2.1

Next: Part 2A-2.2 (Questions 31–40)

CBSE Class 11 Chemistry

Chapter 1: Some Basic Concepts of Chemistry (Mole Concept)

Part 2A-2.2 (Short Answer Questions 31–40)

Q31. Calculate the number of moles present in 9 g of water.

Answer:
Molar mass of H₂O = 18 g/mol
Moles = Mass ÷ Molar Mass = 9 ÷ 18 = 0.5 mol

---

Q32. How many molecules are present in 1 mole of carbon dioxide?

Answer:
1 mole contains Avogadro number of molecules.

Number of molecules = 6.022 × 10²³

---

Q33. Why is the mole concept important in chemistry?

Answer:

• Helps count tiny particles.
• Converts mass into number of particles.
• Used in chemical calculations.
• Essential in stoichiometry.

---

Q34. What is meant by molar volume?

Answer:
The volume occupied by one mole of any gas at STP is called molar volume.

Molar Volume = 22.4 L

---

Q35. Find the molar mass of ammonia (NH₃).

Answer:
N = 14
H = 1 × 3 = 3

Molar mass = 14 + 3 = 17 g/mol

---

Q36. State Avogadro's Law.

Answer:
Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

---

Q37. What is the relationship between mass and moles?

Answer:
Moles = Mass ÷ Molar Mass

or

Mass = Moles × Molar Mass

---

Q38. Calculate the mass of 2 moles of oxygen gas (O₂).

Answer:
Molar mass of O₂ = 32 g/mol

Mass = 2 × 32 = 64 g

---

Q39. What is the value of one Faraday?

Answer:
One Faraday is the charge carried by one mole of electrons.

1 F = 96500 Coulomb

---

Q40. Write any four important formulae of the Mole Concept.

Answer:

1. Moles = Mass ÷ Molar Mass
2. Mass = Moles × Molar Mass
3. Particles = Moles × 6.022 × 10²³
4. Moles = Volume ÷ 22.4 (at STP)

End of Part 2A-2.2

CBSE Class 11 Chemistry Question Bank
Chapter 1 – Some Basic Concepts of Chemistry

CBSE Class 11 Chemistry

Part 2B - Long Answer Questions (5 Marks)

Question 1

Explain the Mole Concept. Why is it important in Chemistry?

Answer:

The mole is the SI unit used to express the amount of substance.

One mole contains 6.022 × 10²³ particles. This number is called Avogadro's Number.

Importance:

• Helps count atoms and molecules.
• Converts mass into number of particles.
• Used in chemical calculations.
• Used in balancing chemical equations.
• Used in stoichiometry.

Question 2

Explain Avogadro's Law with suitable examples.

Answer:

Avogadro's Law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

Formula

V ∝ n

or

V₁/n₁ = V₂/n₂

Examples

• 22.4 L of Hydrogen contains the same number of molecules as 22.4 L of Oxygen at STP.
• One mole of every gas occupies 22.4 L at STP.

Question 3

Write all important formulae of Mole Concept.

Quantity	Formula
Moles	Mass ÷ Molar Mass
Mass	Moles × Molar Mass
Particles	Moles × 6.022 × 10²³
Moles from Particles	Particles ÷ 6.022 × 10²³
Gas Volume	Moles × 22.4 L
Moles from Volume	Volume ÷ 22.4
Charge	Moles × 96500 C

Question 4

Differentiate between Atomic Mass, Molecular Mass and Molar Mass.

Atomic Mass	Molecular Mass	Molar Mass
Mass of one atom.	Mass of one molecule.	Mass of one mole.
Unit = u	Unit = u	Unit = g/mol
Example: C = 12 u	H₂O = 18 u	H₂O = 18 g/mol

Question 5

Explain Molar Mass with examples.

Molar mass is the mass of one mole of any substance.

Substance	Molar Mass
Hydrogen	1 g/mol
Carbon	12 g/mol
Oxygen	16 g/mol
Water	18 g/mol
Carbon dioxide	44 g/mol

Question 6

Explain the relationship between Mole, Mass and Number of Particles.

Mass = Moles × Molar Mass

Moles = Mass ÷ Molar Mass

Particles = Moles × Avogadro Number

Thus, mass can be converted into moles and moles into particles.

Question 7

Calculate the number of molecules present in 36 g of water.

Mass = 36 g

Molar Mass = 18 g/mol

Moles = 36 ÷ 18 = 2 mol

Number of molecules

= 2 × 6.022 × 10²³

= 1.204 × 10²⁴ molecules

Question 8

Calculate the volume occupied by 4 moles of oxygen gas at STP.

Volume = Moles × 22.4

= 4 × 22.4

= 89.6 L

Question 9

Explain Faraday Constant.

• Faraday is the charge carried by one mole of electrons.
• Value = 96500 Coulomb.
• Used in electrochemistry.
• Represented by F.

Question 10

Write five applications of Mole Concept.

• Calculation of mass.
• Calculation of number of atoms.
• Calculation of molecules.
• Calculation of gas volume.
• Chemical equation calculations.
• Industrial chemical production.
• Electrochemistry calculations.

---

Important Formula Revision

• Moles = Mass ÷ Molar Mass
• Mass = Moles × Molar Mass
• Particles = Moles × 6.022 × 10²³
• Moles = Particles ÷ 6.022 × 10²³
• Volume = Moles × 22.4 L
• Moles = Volume ÷ 22.4
• Charge = Moles × 96500 C
• 1 Faraday = 96500 C

CBSE Class 11 Chemistry

Part 2C-1A

Numerical Questions (1–5)

Useful Formulae

Moles = Mass ÷ Molar Mass
Mass = Moles × Molar Mass
Particles = Moles × 6.022 × 10²³
Volume at STP = Moles × 22.4 L

Question 1

Calculate the number of moles present in 36 g of water.

Given:

• Mass = 36 g
• Molar Mass of H₂O = 18 g/mol

Formula:

Moles = Mass ÷ Molar Mass

= 36 ÷ 18

= 2 mol

Answer = 2 moles

Question 2

Calculate the mass of 3 moles of carbon dioxide (CO₂).

Given:

• Moles = 3
• Molar Mass of CO₂ = 44 g/mol

Formula:

Mass = Moles × Molar Mass

= 3 × 44

= 132 g

Answer = 132 g

Question 3

Find the number of molecules present in 2 moles of oxygen gas.

Formula:

Number of Molecules = Moles × 6.022 × 10²³

= 2 × 6.022 × 10²³

= 1.2044 × 10²⁴ molecules

Answer = 1.204 × 10²⁴ molecules

Question 4

Calculate the volume occupied by 4 moles of oxygen gas at STP.

Formula:

Volume = Moles × 22.4

= 4 × 22.4

= 89.6 L

Answer = 89.6 L

Question 5

Calculate the number of atoms present in 1 mole of carbon.

Formula:

Number of Atoms = 1 × 6.022 × 10²³

= 6.022 × 10²³ atoms

Answer = 6.022 × 10²³ atoms

Quick Formula Revision

Quantity	Formula
Moles	Mass ÷ Molar Mass
Mass	Moles × Molar Mass
Particles	Moles × 6.022 × 10²³
Volume (STP)	Moles × 22.4 L

End of Part 2C-1A

Next: Part 2C-1B (Numerical Questions 6–10)

CBSE Class 11 Chemistry

Part 2C-1B : Numerical Questions (6–10)

Important Formulae

Moles = Mass ÷ Molar Mass
Mass = Moles × Molar Mass
Particles = Moles × 6.022 × 10²³
Volume at STP = Moles × 22.4 L
Charge = Moles × 96500 C

Question 6

Calculate the mass of 2.5 moles of oxygen gas (O₂).

Solution

Molar mass of O₂ = 32 g/mol

Mass = Moles × Molar Mass

= 2.5 × 32

= 80 g

Answer: 80 g

Question 7

How many moles are present in 90 g of water (H₂O)?

Solution

Mass = 90 g
Molar mass of H₂O = 18 g/mol

Moles = 90 ÷ 18

= 5 mol

Answer: 5 moles

Question 8

Calculate the number of oxygen molecules present in 0.5 mole of oxygen gas.

Solution

Number of molecules = 0.5 × 6.022 × 10²³

= 3.011 × 10²³ molecules

Answer: 3.011 × 10²³ molecules

Question 9

Find the volume occupied by 3 moles of nitrogen gas at STP.

Solution

Volume = Moles × 22.4

= 3 × 22.4

= 67.2 L

Answer: 67.2 litres

Question 10

Calculate the charge carried by 3 moles of electrons.

Solution

Charge = Moles × Faraday Constant

= 3 × 96500

= 289500 C

Answer: 289500 Coulomb

---

Quick Formula Revision

• Moles = Mass ÷ Molar Mass
• Mass = Moles × Molar Mass
• Particles = Moles × 6.022 × 10²³
• Gas Volume at STP = Moles × 22.4 L
• Charge = Moles × 96500 C

CBSE Class 11 Chemistry
Some Basic Concepts of Chemistry
Part 2C-1B (Numericals 6–10)

CBSE Class 11 Chemistry

Part 2C-2 (Numericals 11–20)

Important Formula:
Moles = Mass ÷ Molar Mass
Particles = Moles × 6.022 × 10²³
Volume at STP = Moles × 22.4 L

Question 11

Calculate the number of moles present in 72 g of water.

Molar mass of H₂O = 18 g/mol
Moles = 72 ÷ 18 = 4 mol

---

Question 12

Find the mass of 5 moles of oxygen gas (O₂).

Molar mass of O₂ = 32 g/mol
Mass = 5 × 32 = 160 g

---

Question 13

Calculate the number of molecules present in 4 moles of carbon dioxide.

Number of molecules = 4 × 6.022 × 10²³
=
2.4088 × 10²⁴ molecules

---

Question 14

Find the volume occupied by 5 moles of nitrogen gas at STP.

Volume = 5 × 22.4
=
112 L

---

Question 15

Calculate the number of atoms present in 3 moles of helium.

Atoms = 3 × 6.022 × 10²³
=
1.8066 × 10²⁴ atoms

---

Question 16

Calculate the mass of 0.75 mole of sodium chloride.

Molar mass of NaCl = 58.5 g/mol
Mass = 0.75 × 58.5
=
43.875 g

---

Question 17

Find the number of moles in 98 g of sulphuric acid (H₂SO₄).

Molar mass = 98 g/mol
Moles = 98 ÷ 98
=
1 mole

---

Question 18

Calculate the number of molecules in 0.25 mole of methane.

Number of molecules = 0.25 × 6.022 × 10²³
=
1.5055 × 10²³ molecules

---

Question 19

Find the volume occupied by 0.5 mole of hydrogen gas at STP.

Volume = 0.5 × 22.4
=
11.2 L

---

Question 20

Calculate the mass of 2.5 moles of carbon dioxide.

Molar mass of CO₂ = 44 g/mol
Mass = 2.5 × 44
=
110 g

---

Quick Formula Revision

• Moles = Mass ÷ Molar Mass
• Mass = Moles × Molar Mass
• Particles = Moles × 6.022 × 10²³
• Volume = Moles × 22.4 L (STP)
• 1 Mole = 6.022 × 10²³ particles
• 1 Mole Gas = 22.4 L at STP

CBSE Class 11 Chemistry

Part 2C-3A

Numerical Problems (Questions 21–25)

Question 21

Calculate the number of moles present in 72 g of water (H₂O).

Formula: Moles = Mass ÷ Molar Mass

Mass = 72 g
Molar Mass of H₂O = 18 g/mol

Moles = 72 ÷ 18
= 4 mol

Answer: 4 moles

---

Question 22

Calculate the number of molecules present in 4 moles of carbon dioxide.

Formula: Number of Molecules = Moles × Avogadro Number

Number of Molecules
= 4 × 6.022 × 10²³
= 2.4088 × 10²⁴ molecules

Answer: 2.4088 × 10²⁴ molecules

---

Question 23

Find the mass of 3 moles of sodium chloride (NaCl).

Formula: Mass = Moles × Molar Mass

Molar Mass of NaCl
= 23 + 35.5
= 58.5 g/mol

Mass
= 3 × 58.5
= 175.5 g

Answer: 175.5 g

---

Question 24

Calculate the volume occupied by 2.5 moles of nitrogen gas at STP.

Formula: Volume = Moles × 22.4 L

Volume
= 2.5 × 22.4
= 56.0 L

Answer: 56.0 L

---

Question 25

Calculate the charge carried by 3 moles of electrons.

Formula: Charge = Moles × Faraday Constant

Faraday Constant = 96500 C/mol

Charge
= 3 × 96500
= 289500 C

Answer: 289500 C

---

Quick Formula Revision

Moles = Mass ÷ Molar Mass

Mass = Moles × Molar Mass

Particles = Moles × 6.022 × 10²³

Moles = Particles ÷ 6.022 × 10²³

Volume at STP = Moles × 22.4 L

Charge = Moles × 96500 C

CBSE Class 11 Chemistry

Part 2C-3B : Numerical Questions (26–30)

Important Formulae

Moles = Mass ÷ Molar Mass
Mass = Moles × Molar Mass
Particles = Moles × 6.022 × 10²³
Volume at STP = Moles × 22.4 L

Question 26

Calculate the mass of 2.5 moles of oxygen gas (O₂).

Solution Molar mass of O₂ = 32 g/mol Mass = Moles × Molar Mass = 2.5 × 32 = 80 g

Answer: 80 g

Question 27

Calculate the number of moles present in 49 g of sulphuric acid (H₂SO₄).

Solution Molar Mass of H₂SO₄ = 2 + 32 + 64 = 98 g/mol Moles = Mass ÷ Molar Mass = 49 ÷ 98 = 0.5 mol

Answer: 0.5 mol

Question 28

Find the number of molecules present in 3 moles of ammonia (NH₃).

Solution Number of molecules = 3 × 6.022 × 10²³ = 18.066 × 10²³ = 1.8066 × 10²⁴ molecules

Answer: 1.8066 × 10²⁴ molecules

Question 29

Calculate the volume occupied by 1.5 moles of nitrogen gas at STP.

Solution Volume = Moles × 22.4 = 1.5 × 22.4 = 33.6 L

Answer: 33.6 L

Question 30

How many atoms are present in 2 moles of helium?

Solution 1 mole Helium = 6.022 × 10²³ atoms 2 moles = 2 × 6.022 × 10²³ = 12.044 × 10²³ = 1.2044 × 10²⁴ atoms

Answer: 1.2044 × 10²⁴ atoms

---

Quick Revision

• 1 Mole = 6.022 × 10²³ particles
• 1 Mole Gas at STP = 22.4 L
• Moles = Mass ÷ Molar Mass
• Mass = Moles × Molar Mass
• Particles = Moles × Avogadro Number
• Volume = Moles × 22.4 L

End of Part 2C-3B

CBSE Class 11 Chemistry – Some Basic Concepts of Chemistry

CBSE Class 11 Chemistry

Part 2C-4 Numerical Questions (31–35)

Important Formulae

Moles = Mass ÷ Molar Mass
Particles = Moles × 6.022 × 10²³
Volume at STP = Moles × 22.4 L
Charge = Moles × 96500 C

Question 31

Calculate the number of moles present in 98 g of H₂SO₄.

Solution Atomic masses: H = 1 S = 32 O = 16

Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 98 g/mol

Moles = Mass ÷ Molar Mass
= 98 ÷ 98
= 1 mol

Answer: 1 mole

---

Question 32

Calculate the mass of 3 moles of carbon dioxide (CO₂).

Solution Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol

Mass = Moles × Molar Mass
= 3 × 44
= 132 g

Answer: 132 g

---

Question 33

Calculate the number of molecules present in 0.5 mole of oxygen (O₂).

Solution Number of molecules = Moles × Avogadro Number
= 0.5 × 6.022 × 10²³
= 3.011 × 10²³ molecules

Answer: 3.011 × 10²³ molecules

---

Question 34

Calculate the volume occupied by 2.5 moles of nitrogen gas at STP.

Solution Volume = Moles × 22.4
= 2.5 × 22.4
= 56.0 L

Answer: 56 L

---

Question 35

Calculate the charge carried by 4 moles of electrons.

Solution Charge = Moles × Faraday Constant
= 4 × 96500
= 386000 Coulomb

Answer: 386000 C

---

Quick Revision

• Avogadro Number = 6.022 × 10²³ particles/mol
• 1 Mole Gas at STP = 22.4 L
• 1 Faraday = 96500 C
• Moles = Mass ÷ Molar Mass
• Mass = Moles × Molar Mass
• Particles = Moles × 6.022 × 10²³
• Volume = Moles × 22.4 L
• Charge = Moles × 96500 C

CBSE Class 11 Chemistry

Part 2C-4 Numerical Problems (Questions 36–40)

Question 36

Calculate the number of moles present in 90 g of glucose (C₆H₁₂O₆).

Given:

Mass = 90 g

Molar Mass of C₆H₁₂O₆

= (6 × 12) + (12 × 1) + (6 × 16)

= 72 + 12 + 96

= 180 g/mol

Moles = Mass ÷ Molar Mass

Moles = 90 ÷ 180

= 0.5 mol

Answer: 0.5 mole

---

Question 37

Calculate the mass of 3 moles of carbon dioxide (CO₂).

Given:

Moles = 3

Molar Mass of CO₂ = 44 g/mol

Mass = Moles × Molar Mass

Mass = 3 × 44

= 132 g

Answer: 132 g

---

Question 38

Calculate the volume occupied by 5 moles of nitrogen gas at STP.

Given:

Moles = 5

1 mole gas occupies 22.4 L at STP.

Volume = Moles × 22.4 L

Volume = 5 × 22.4

= 112 L

Answer: 112 L

---

Question 39

Calculate the number of molecules present in 0.25 mole of ammonia (NH₃).

Given:

Moles = 0.25

Number of Molecules = Moles × 6.022 × 10²³

= 0.25 × 6.022 × 10²³

= 1.5055 × 10²³ molecules

Answer: 1.5055 × 10²³ molecules

---

Question 40

Calculate the charge carried by 4 moles of electrons.

Given:

Moles = 4

1 mole of electrons carries 96500 C.

Charge = Moles × 96500 C

Charge = 4 × 96500

= 386000 C

Answer: 386000 C

---

Important Formulae Used

• Moles = Mass ÷ Molar Mass
• Mass = Moles × Molar Mass
• Volume at STP = Moles × 22.4 L
• Number of Molecules = Moles × 6.022 × 10²³
• Charge = Moles × 96500 C

Quick Answers

Question	Answer
36	0.5 mol
37	132 g
38	112 L
39	1.5055 × 10²³ molecules
40	386000 C

CBSE Class 11 Chemistry

Part 2D-1A

Assertion–Reason Questions (1–15)

Directions: Choose the correct option.

A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. Both Assertion and Reason are true but Reason is not the correct explanation.
C. Assertion is true but Reason is false.
D. Assertion is false but Reason is true.

Question 1

Assertion (A): One mole of every substance contains 6.022 × 10²³ particles.

Reason (R): Avogadro number represents the number of particles present in one mole.

Answer: A

Question 2

Assertion (A): One mole of oxygen molecules contains 6.022 × 10²³ molecules.

Reason (R): Every mole contains Avogadro number of particles.

Answer: A

Question 3

Assertion (A): One mole of carbon atoms weighs 12 g.

Reason (R): Atomic mass of carbon is 12 u.

Answer: A

Question 4

Assertion (A): Molar mass is expressed in g/mol.

Reason (R): Molar mass is the mass of one mole of a substance.

Answer: A

Question 5

Assertion (A): One mole of gas occupies 22.4 L at STP.

Reason (R): Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Answer: A

Question 6

Assertion (A): Moles can be calculated using mass and molar mass.

Reason (R): Moles = Mass ÷ Molar Mass.

Answer: A

Question 7

Assertion (A): Number of particles increases as number of moles increases.

Reason (R): Particles = Moles × Avogadro Number.

Answer: A

Question 8

Assertion (A): One Faraday is equal to 96500 C.

Reason (R): One mole of electrons carries a charge of 96500 C.

Answer: A

Question 9

Assertion (A): Water has a molar mass of 18 g/mol.

Reason (R): Water contains two hydrogen atoms and one oxygen atom.

Answer: A

Question 10

Assertion (A): One mole of hydrogen gas and one mole of oxygen gas contain the same number of molecules.

Reason (R): Every mole contains Avogadro number of particles.

Answer: A

Question 11

Assertion (A): Molar mass of NaCl is 58.5 g/mol.

Reason (R): Sodium has atomic mass 23 u and chlorine has atomic mass 35.5 u.

Answer: A

Question 12

Assertion (A): Mole is the SI unit of amount of substance.

Reason (R): SI system uses mole to measure the quantity of particles.

Answer: A

Question 13

Assertion (A): Carbon dioxide has a molar mass of 44 g/mol.

Reason (R): Carbon has atomic mass 12 u and oxygen has atomic mass 16 u.

Answer: A

Question 14

Assertion (A): One mole of methane contains 6.022 × 10²³ methane molecules.

Reason (R): Methane is a molecular compound.

Answer: B

Question 15

Assertion (A): Atomic mass and molar mass have the same numerical value but different units.

Reason (R): Atomic mass is expressed in u, while molar mass is expressed in g/mol.

Answer: A

---

End of Part 2D-1A

Next File: Part 2D-1B (Assertion–Reason Questions 16–30)

CBSE Class 11 Chemistry

Assertion–Reason Questions (16–30)

Choose the correct option:

1. Both Assertion and Reason are true, and Reason is the correct explanation.
2. Both Assertion and Reason are true, but Reason is not the correct explanation.
3. Assertion is true, but Reason is false.
4. Assertion is false, but Reason is true.

16.

Assertion: One mole of sodium contains 6.022 × 10²³ atoms.
Reason: One mole always contains Avogadro's number of particles.

Answer: 1

17.

Assertion: Molar mass of CO₂ is 44 g/mol.
Reason: Carbon has atomic mass 12 u and oxygen has atomic mass 16 u.

Answer: 1

18.

Assertion: One mole of oxygen atoms has a mass of 16 g.
Reason: Atomic mass of oxygen is 16 u.

Answer: 1

19.

Assertion: Number of molecules in one mole of water equals Avogadro's number.
Reason: Every mole contains the same number of elementary entities.

Answer: 1

20.

Assertion: At STP, one mole of every gas occupies 22.4 L.
Reason: Gas volume depends only on pressure.

Answer: 3

21.

Assertion: Molar mass is expressed in g/mol.
Reason: Atomic mass is expressed in grams.

Answer: 3

22.

Assertion: One Faraday equals 96500 C.
Reason: One mole of electrons carries 96500 C charge.

Answer: 1

23.

Assertion: One mole of hydrogen molecules contains twice the number of hydrogen atoms.
Reason: Each H₂ molecule contains two hydrogen atoms.

Answer: 1

24.

Assertion: One mole of NaCl contains formula units.
Reason: NaCl is an ionic compound.

Answer: 1

25.

Assertion: Number of moles increases when mass increases for the same substance.
Reason: Moles = Mass ÷ Molar Mass.

Answer: 1

26.

Assertion: Avogadro number has unit mol⁻¹.
Reason: It represents particles present in one mole.

Answer: 1

27.

Assertion: Molecular mass of NH₃ is 17 u.
Reason: Nitrogen has atomic mass 14 u and hydrogen has atomic mass 1 u.

Answer: 1

28.

Assertion: Equal masses of different substances always contain equal number of molecules.
Reason: Number of molecules depends upon molar mass.

Answer: 4

29.

Assertion: Mole concept is useful in stoichiometric calculations.
Reason: Chemical reactions occur according to mole ratios.

Answer: 1

30.

Assertion: The SI unit of amount of substance is mole.
Reason: Mole helps express large numbers of atoms and molecules conveniently.

Answer: 2

---

Answer Key

Q.No.	Answer	Q.No.	Answer
16	1	24	1
17	1	25	1
18	1	26	1
19	1	27	1
20	3	28	4
21	3	29	1
22	1	30	2
23	1	-	-

End of Part 2D-1B
CBSE Class 11 Chemistry – Some Basic Concepts of Chemistry (Mole Concept)

CBSE Class 11 Chemistry

Part 2D-2A : Statement Based Questions (1–15)

Directions: Read Statement I and Statement II carefully and choose the correct option.

Options:

A. Both Statement I and Statement II are true, and Statement II is the correct explanation of Statement I.

B. Both Statement I and Statement II are true, but Statement II is NOT the correct explanation of Statement I.

C. Statement I is true, but Statement II is false.

D. Statement I is false, but Statement II is true.

---

Question 1

Statement I: One mole of every substance contains 6.022 × 10²³ particles.
Statement II: This number is known as Avogadro's number.

Answer: A

Question 2

Statement I: One mole of oxygen gas contains the same number of molecules as one mole of hydrogen gas.
Statement II: Every mole contains Avogadro's number of particles.

Answer: A

Question 3

Statement I: Molar mass is expressed in g/mol.
Statement II: Molar mass is equal to molecular mass expressed in grams.

Answer: A

Question 4

Statement I: The SI unit of amount of substance is mole.
Statement II: Mole is used for counting microscopic particles.

Answer: A

Question 5

Statement I: One mole of gas occupies 22.4 L at STP.
Statement II: STP means 273 K temperature and 1 atm pressure.

Answer: B

Question 6

Statement I: Carbon has atomic mass 12 u.
Statement II: One mole of carbon weighs 12 g.

Answer: B

Question 7

Statement I: Number of moles increases with increase in mass.
Statement II: Moles = Mass ÷ Molar Mass.

Answer: A

Question 8

Statement I: One Faraday equals 96500 coulomb.
Statement II: One mole of electrons carries 96500 coulomb charge.

Answer: A

Question 9

Statement I: Water has molar mass 18 g/mol.
Statement II: Water contains two hydrogen atoms and one oxygen atom.

Answer: A

Question 10

Statement I: Molar mass of CO₂ is 44 g/mol.
Statement II: Carbon dioxide contains one carbon atom and two oxygen atoms.

Answer: A

Question 11

Statement I: One mole of NaCl contains Avogadro's number of formula units.
Statement II: NaCl is an ionic compound.

Answer: A

Question 12

Statement I: Mole concept helps in stoichiometric calculations.
Statement II: It relates mass, particles and volume.

Answer: A

Question 13

Statement I: Molecular mass is expressed in atomic mass unit (u).
Statement II: Molar mass is expressed in g/mol.

Answer: B

Question 14

Statement I: One mole of methane contains one mole of carbon atoms.
Statement II: Each methane molecule contains one carbon atom.

Answer: A

Question 15

Statement I: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Statement II: This statement is known as Avogadro's Law.

Answer: A

---

Answer Key

Question	Answer
1	A
2	A
3	A
4	A
5	B
6	B
7	A
8	A
9	A
10	A
11	A
12	A
13	B
14	A
15	A

CBSE Class 11 Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept)
Part 2D-2A – Statement Based Questions (1–15)

CBSE Class 11 Chemistry

Part 2D-2B: Statement Based Questions (16–30)

Q16.
Statement I: One mole of NaCl contains Avogadro number of formula units.
Statement II: NaCl is an ionic compound.

Choose the correct option:

A. Both statements are true and Statement II correctly explains Statement I.
B. Both statements are true but Statement II does not explain Statement I.
C. Statement I is true but Statement II is false.
D. Statement I is false but Statement II is true.

Answer: A

Q17.
Statement I: Molar mass of CO₂ is 44 g/mol.
Statement II: CO₂ contains one carbon atom and two oxygen atoms.

Answer: A

Q18.
Statement I: Atomic mass is expressed in unified atomic mass unit (u).
Statement II: Molar mass is expressed in g/mol.

Answer: B

Q19.
Statement I: One mole of H₂O contains 6.022 × 10²³ molecules.
Statement II: Every water molecule contains three atoms.

Answer: B

Q20.
Statement I: Mole is the SI unit of amount of substance.
Statement II: Mole is used to count microscopic particles.

Answer: A

Q21.
Statement I: Moles = Mass ÷ Molar Mass.
Statement II: Molar mass is expressed in grams per mole.

Answer: A

Q22.
Statement I: One mole of O₂ contains twice the number of oxygen atoms as one mole of O atoms.
Statement II: One O₂ molecule contains two oxygen atoms.

Answer: A

Q23.
Statement I: One mole of electrons carries one Faraday of charge.
Statement II: One Faraday equals 96500 C.

Answer: A

Q24.
Statement I: Relative atomic mass has no unit.
Statement II: It is measured relative to 1/12th mass of carbon-12 atom.

Answer: A

Q25.
Statement I: Molar volume is applicable to gases at STP.
Statement II: One mole of every gas occupies 22.4 L at STP.

Answer: A

Q26.
Statement I: Molecular mass of NH₃ is 17 u.
Statement II: NH₃ contains one nitrogen atom and three hydrogen atoms.

Answer: A

Q27.
Statement I: One mole of carbon atoms weighs 12 g.
Statement II: Atomic mass of carbon is 12 u.

Answer: A

Q28.
Statement I: Number of particles = Moles × Avogadro Number.
Statement II: Avogadro Number equals 6.022 × 10²³.

Answer: A

Q29.
Statement I: One mole of CO₂ contains one mole of carbon atoms.
Statement II: Every CO₂ molecule contains one carbon atom.

Answer: A

Q30.
Statement I: Mole concept is useful in stoichiometric calculations.
Statement II: It helps convert mass into moles and particles.

Answer: A

---

Answer Key

Question	Answer	Question	Answer
16	A	24	A
17	A	25	A
18	B	26	A
19	B	27	A
20	A	28	A
21	A	29	A
22	A	30	A
23	A

End of Part 2D-2B

CBSE Class 11 Chemistry | Some Basic Concepts of Chemistry | Statement-Based Questions (16–30)

CBSE Class 11 Chemistry

Mole Concept

PART 3A

Case Study Questions

Case Study 1

Ravi measured 18 g of water. He wanted to calculate the number of moles and molecules present in it. The molar mass of water is 18 g/mol. Use the information to answer the following questions.

Q1. What is the molar mass of water?
Answer: 18 g/mol

Q2. Number of moles present?
Answer: 1 mole

Q3. Number of molecules?
Answer: 6.022 × 10²³ molecules

Q4. Number of hydrogen atoms?
Answer: 1.204 × 10²⁴ atoms

Q5. Number of oxygen atoms?
Answer: 6.022 × 10²³ atoms

Case Study 2

A cylinder contains 2 moles of oxygen gas at STP. Use the information to answer the questions.

Q1. Volume occupied?
Answer: 44.8 L

Q2. Number of molecules?
Answer: 1.204 × 10²⁴ molecules

Q3. Number of oxygen atoms?
Answer: 2.408 × 10²⁴ atoms

Q4. Molar mass of O₂?
Answer: 32 g/mol

Q5. Mass of oxygen gas?
Answer: 64 g

Case Study 3

Neha has 44 g of carbon dioxide (CO₂). Answer the following questions.

Q1. Molar mass of CO₂?
Answer: 44 g/mol

Q2. Number of moles?
Answer: 1 mole

Q3. Number of molecules?
Answer: 6.022 × 10²³ molecules

Q4. Number of oxygen atoms?
Answer: 1.204 × 10²⁴ atoms

Q5. Number of carbon atoms?
Answer: 6.022 × 10²³ atoms

Match the Columns

Column A	Column B
1. Mole	a. 96500 C
2. Avogadro Number	b. SI Unit
3. Faraday	c. 22.4 L
4. STP Volume	d. 6.022 × 10²³
5. Molar Mass	e. g/mol

Answers

Question	Answer
1	b
2	d
3	a
4	c
5	e

Match the Columns - Set 2

Column A	Column B
Hydrogen	1 g/mol
Carbon	12 g/mol
Oxygen	16 g/mol
Water	18 g/mol
Carbon Dioxide	44 g/mol

Answers

Substance	Correct Match
Hydrogen	1 g/mol
Carbon	12 g/mol
Oxygen	16 g/mol
Water	18 g/mol
Carbon Dioxide	44 g/mol

Match the Columns - Set 3

Column A	Column B
Mass	Mole × Molar Mass
Moles	Mass ÷ Molar Mass
Particles	Mole × Avogadro Number
Gas Volume	Mole × 22.4
Charge	Mole × 96500

Answers

Column A	Correct Match
Mass	Mole × Molar Mass
Moles	Mass ÷ Molar Mass
Particles	Mole × Avogadro Number
Gas Volume	Mole × 22.4
Charge	Mole × 96500

End of Part 3A

CBSE Class 11 Chemistry | Mole Concept

Next: Part 3B - HOTS Questions, Competency-Based Questions & Previous Year Questions

CBSE Class 11 Chemistry | Mole Concept | Part 3B

CBSE Class 11 Chemistry

Part 3B : HOTS Questions + Competency Based Questions + Previous Year Questions

Section A : HOTS (Higher Order Thinking Skills)

Q1. One mole of hydrogen gas and one mole of oxygen gas have equal number of molecules but different masses. Explain why.

Answer:
Both contain Avogadro number (6.022 × 10²³) of molecules because each is one mole. Mass differs because H₂ = 2 g/mol O₂ = 32 g/mol Hence number of molecules is same but masses are different.

Q2. Which contains more atoms? 18 g water or 44 g carbon dioxide?

Answer: 18 g H₂O = 1 mole Each molecule has 3 atoms Total atoms = 3 × Avogadro Number 44 g CO₂ = 1 mole Each molecule has 3 atoms Total atoms = 3 × Avogadro Number Therefore both contain equal number of atoms.

Q3. Can two substances having different masses contain equal number of molecules?

Answer: Yes. Example 18 g H₂O 44 g CO₂ Both are one mole and contain equal number of molecules.

Q4. Why is mole called a counting unit?

Answer: Atoms are extremely small. Instead of counting atoms individually, chemists count them in groups of 6.022 × 10²³ called one mole.

Q5. A gas occupies 44.8 L at STP. Calculate number of moles.

Answer: Moles = Volume / 22.4 = 44.8 / 22.4 = 2 mol

---

Section B : Competency Based Questions

Q1. A student weighs 36 g of water. Answer the following.

1. Find number of moles.
2. Find number of molecules.
3. Find total hydrogen atoms.

Moles = 36/18 = 2 Molecules = 2 × 6.022 ×10²³ = 1.204 ×10²⁴ Hydrogen atoms = 2 × molecules = 2.408 ×10²⁴ atoms

Q2. Rahul takes one mole of sodium chloride. Answer:

1. Number of formula units?
2. Mass?
3. Molar mass?

Formula units = 6.022 ×10²³ Mass = 58.5 g Molar mass = 58.5 g/mol

Q3. One mole of electrons passes through a conductor. Find

1. Total charge
2. Name of this quantity

Charge = 96500 C Quantity = One Faraday

Q4. A balloon contains 3 moles of helium gas. Calculate volume at STP.

Volume = 3 × 22.4 = 67.2 L

Q5. A sample contains 12.044 ×10²³ molecules. Calculate moles.

Moles = 12.044 ×10²³ ÷ 6.022 ×10²³ = 2 mol

---

Section C : Previous Year CBSE Questions (Practice)

PYQ 1. Define mole.

One mole is the amount of substance containing 6.022 ×10²³ particles.

PYQ 2. Calculate number of moles in 22 g carbon dioxide.

Moles = 22/44 = 0.5 mol

PYQ 3. Calculate mass of 0.25 mole oxygen gas.

Mass = 0.25 ×32 = 8 g

PYQ 4. State Avogadro Law.

Equal volumes of gases at same temperature and pressure contain equal number of molecules.

PYQ 5. Calculate molecules present in 3 moles methane.

3 × 6.022 ×10²³ = 1.806 ×10²⁴ molecules

PYQ 6. Calculate volume occupied by 4 moles nitrogen gas at STP.

Volume = 4 ×22.4 = 89.6 L

PYQ 7. What is molar mass?

Mass of one mole of a substance is called molar mass.

PYQ 8. Find charge carried by 2 moles of electrons.

Charge = 2 ×96500 = 193000 C

---

Important Formula Sheet

Formula	Expression
Moles	Mass ÷ Molar Mass
Mass	Moles × Molar Mass
Particles	Moles × 6.022 ×10²³
Moles	Particles ÷ 6.022 ×10²³
Gas Volume	Moles ×22.4 L
Charge	Moles ×96500 C

End of Part 3B

CBSE Class 11 Chemistry
Chapter : Some Basic Concepts of Chemistry
Mole Concept

Part 3C: Formula Sheet + Chapter Summary + Complete Answer Key + Print-Friendly HTML Footer