Chemistry
Mole Concept
Workbook
Type-wise Discussion
and Step by Step Solved
Examples
User
, Mole concept on Number of Molecules, Mass in gram and Volume
1
🌿 Preface
This document offers a complete and concept-based exploration of the Mole Concept and
Molecular Mass Determination. It is designed to help learners understand how to relate
mass, moles, molecules, and volume through clear, logical explanations and problem-solving
techniques.
Each topic is organized type-wise and supported with step-by-step methods, formulas, and
solved examples. From determining the number of particles to calculating molecular mass or
volume, every concept is explained in a simple and structured way that builds deep
understanding.
Special attention has been given to breaking complex numerical problems into smaller,
manageable parts so that students can easily follow each calculation and reasoning. The
inclusion of MCQ practice exercises at the end of every type allows learners to test their
understanding and improve exam readiness.
Whether you are a school or college student, this sheet serves as a complete study guide for
mastering the mole concept—making chemistry calculations easier, clearer, and more
practical for real exam success.
, Mole concept on Number of Molecules, Mass in gram and Volume
2
Table of contents with Formulae
Type 👉 Topics 👉 Formula Page
1 Determining the number of N w 04
=
molecules/atoms when the mass is 6.023 × 1023 M
given.
Mole is given — find the number of N = n × 6.023 × 1023
molecules. 09
2(a)
Find the number of atoms in molecules. N = n × 6.023 × 1023 ×
number of atoms 12
2(b)
Find the number of molecules when (Formula: N = n × 6.023 ×
Number of ions is given. 1023 ) ; where n = number of
02(c) 17
moles of that ion.
Volume is given — find the number of
3 N 𝑉 21
molecules. = 22.4
6.023×1023
Number is given — find the mass in w
=
N
4 gram. M 6.023 × 1023 24
Volume is given — find the mass in
w V
5 gram. M
= 22.4 27
Mole is given — find the mass in 𝑤 = 𝑛𝑀
6 gram. 29
Number is given — find the volume. 𝑉
=
N
7 22.4 6.023 × 1023 30
Mass is given in grams — find the 𝑉
=
w
8 volume. 22.4 M 35
9 Mole is given — find the volume. 𝑉 = 𝑛 × 22.4 39
N
𝑛=
10 Number is given — find the mole. 6.023 × 1023 41
w
11 𝑛=M
Mass is given — find the mole. 43
12 𝑉
𝑛 = 22.4
Volume is given — find the mole. 44
, Mole concept on Number of Molecules, Mass in gram and Volume
3
👉 The Process of Determining the atoms, molecules and ions number
from mass in gram and Molecular Mass of a Molecule:
Avogadro’s Hypothesis:
Statement: “At the same temperature and pressure, equal volumes of all gases contain an
equal number of molecules.”
This means the type of gas does not matter — hydrogen, oxygen, nitrogen, carbon dioxide,
SO2 or any other gas — if their volumes are the same and the conditions of temperature and
pressure are the same, then the number of molecules inside will also be the same.
Mathematical Form:
𝐕∝𝐧
Where (Temperature and Pressure constant)
• V = Volume of gas
• n = Number of moles
So, 1 mole of any gas at STP (Standard Temperature and Pressure: 0°C and 1 atm) occupies
22.4 L.
Example:
• 1 L of 𝐍₂ gas at STP and 1 L of 𝐂𝐎₂ gas at STP contain the same number of
molecules, even though oxygen is heavier than hydrogen.
👉 Specifically, if each contains 1 mole (22.4 L), then both have 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 molecules
Application:
Avogadro’s Hypothesis is very useful for:
• Explaining gas reactions (using volume ratios directly instead of mass).
• For determining molar mass of gases.
• Deriving the Ideal Gas Equation: 𝐏𝐕 = 𝐧𝐑𝐓
Mole: The word mole is the unit of any chemical measurement. Suppose,
12 of H2 = 1 dozen ofH2
100 of H2 = 1 hundred of H2
1000 of H2 = 1 thousand ofH2
Similarly, 6.023 × 1023 nos H2 = 1 moleH2
1 mole atom contains 6.023 × 1023 atoms
1 mole molecule contains 6.023 × 1023 molecules
1 mole ion contains 6.023 × 1023 ions
Therefore, the number 6.023 × 1023 is used in the case with atoms, molecules and ions etc.
This number is called the Avogadro number.
Mole Concept
Workbook
Type-wise Discussion
and Step by Step Solved
Examples
User
, Mole concept on Number of Molecules, Mass in gram and Volume
1
🌿 Preface
This document offers a complete and concept-based exploration of the Mole Concept and
Molecular Mass Determination. It is designed to help learners understand how to relate
mass, moles, molecules, and volume through clear, logical explanations and problem-solving
techniques.
Each topic is organized type-wise and supported with step-by-step methods, formulas, and
solved examples. From determining the number of particles to calculating molecular mass or
volume, every concept is explained in a simple and structured way that builds deep
understanding.
Special attention has been given to breaking complex numerical problems into smaller,
manageable parts so that students can easily follow each calculation and reasoning. The
inclusion of MCQ practice exercises at the end of every type allows learners to test their
understanding and improve exam readiness.
Whether you are a school or college student, this sheet serves as a complete study guide for
mastering the mole concept—making chemistry calculations easier, clearer, and more
practical for real exam success.
, Mole concept on Number of Molecules, Mass in gram and Volume
2
Table of contents with Formulae
Type 👉 Topics 👉 Formula Page
1 Determining the number of N w 04
=
molecules/atoms when the mass is 6.023 × 1023 M
given.
Mole is given — find the number of N = n × 6.023 × 1023
molecules. 09
2(a)
Find the number of atoms in molecules. N = n × 6.023 × 1023 ×
number of atoms 12
2(b)
Find the number of molecules when (Formula: N = n × 6.023 ×
Number of ions is given. 1023 ) ; where n = number of
02(c) 17
moles of that ion.
Volume is given — find the number of
3 N 𝑉 21
molecules. = 22.4
6.023×1023
Number is given — find the mass in w
=
N
4 gram. M 6.023 × 1023 24
Volume is given — find the mass in
w V
5 gram. M
= 22.4 27
Mole is given — find the mass in 𝑤 = 𝑛𝑀
6 gram. 29
Number is given — find the volume. 𝑉
=
N
7 22.4 6.023 × 1023 30
Mass is given in grams — find the 𝑉
=
w
8 volume. 22.4 M 35
9 Mole is given — find the volume. 𝑉 = 𝑛 × 22.4 39
N
𝑛=
10 Number is given — find the mole. 6.023 × 1023 41
w
11 𝑛=M
Mass is given — find the mole. 43
12 𝑉
𝑛 = 22.4
Volume is given — find the mole. 44
, Mole concept on Number of Molecules, Mass in gram and Volume
3
👉 The Process of Determining the atoms, molecules and ions number
from mass in gram and Molecular Mass of a Molecule:
Avogadro’s Hypothesis:
Statement: “At the same temperature and pressure, equal volumes of all gases contain an
equal number of molecules.”
This means the type of gas does not matter — hydrogen, oxygen, nitrogen, carbon dioxide,
SO2 or any other gas — if their volumes are the same and the conditions of temperature and
pressure are the same, then the number of molecules inside will also be the same.
Mathematical Form:
𝐕∝𝐧
Where (Temperature and Pressure constant)
• V = Volume of gas
• n = Number of moles
So, 1 mole of any gas at STP (Standard Temperature and Pressure: 0°C and 1 atm) occupies
22.4 L.
Example:
• 1 L of 𝐍₂ gas at STP and 1 L of 𝐂𝐎₂ gas at STP contain the same number of
molecules, even though oxygen is heavier than hydrogen.
👉 Specifically, if each contains 1 mole (22.4 L), then both have 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 molecules
Application:
Avogadro’s Hypothesis is very useful for:
• Explaining gas reactions (using volume ratios directly instead of mass).
• For determining molar mass of gases.
• Deriving the Ideal Gas Equation: 𝐏𝐕 = 𝐧𝐑𝐓
Mole: The word mole is the unit of any chemical measurement. Suppose,
12 of H2 = 1 dozen ofH2
100 of H2 = 1 hundred of H2
1000 of H2 = 1 thousand ofH2
Similarly, 6.023 × 1023 nos H2 = 1 moleH2
1 mole atom contains 6.023 × 1023 atoms
1 mole molecule contains 6.023 × 1023 molecules
1 mole ion contains 6.023 × 1023 ions
Therefore, the number 6.023 × 1023 is used in the case with atoms, molecules and ions etc.
This number is called the Avogadro number.
