IB DP Chemistry - S1.4.1 The mole and Avogadro constant - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S1.4.1 The mole and Avogadro constant – Study Notes – New Syllabus
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Structure 1.4.1 — The Mole and Avogadro Constant
Structure 1.4.1 — The Mole and Avogadro Constant
Definition of a Mole
The mole (mol) is the SI unit of amount of substance.
One mole contains exactly 6.022 × 10²³ elementary entities. This number is called the Avogadro constant, denoted by \( N_A \).
Avogadro Constant:
\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
Elementary entities may be:
- Atoms (e.g. He)
- Molecules (e.g. \( \text{H}_2, \text{H}_2\text{O} \))
- Ions (e.g. \( \text{Na}^+, \text{Cl}^- \))
- Electrons
- Other specified particles
Relationship Between Moles and Number of Particles
The number of particles \( N \) in a sample is related to the amount in moles \( n \) by the formula:
\( N = n \times N_A \)
To find moles from number of particles:
\( n = \frac{N}{N_A} \)
Units:
- \( N \) = number of particles (no unit)
- \( n \) = amount of substance in moles (mol)
- \( N_A \) = Avogadro constant \( = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
Where, m = Given mass, M = Molar mass
v = Given volume, V = Molar volume = 22.4 dm3
n = Number of moles = m/M
Number of atoms = Number of molecules × Atomicity
Important Concept: These calculations assume a sample of pure substance, where each particle is the same kind of entity (e.g. all water molecules).
Example
Calculate the number of water molecules in 0.25 mol of water.
▶️Answer/Explanation
Use the formula:
\( N = n \times N_A \)
\( N = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23} \) molecules
Example
An ion sample contains \( 1.8066 \times 10^{24} \) \( \text{Na}^+ \) ions. Calculate the amount of substance in moles.
▶️Answer/Explanation
Use the formula:
\( n = \frac{N}{N_A} \)
\( n = \frac{1.8066 \times 10^{24}}{6.022 \times 10^{23}} = 3.00 \, \text{mol} \)
Example
How many atoms are present in 0.75 mol of helium gas?
▶️Answer/Explanation
Since helium is monoatomic:
\( N = n \times N_A = 0.75 \times 6.022 \times 10^{23} = 4.517 \times 10^{23} \) atoms