ChemistryICSE Class 10 · Mole Concept and Stoichiometry

Mole, Avogadro Number and Molar Volume: ICSE Class 10 Chemistry

Q: What is the difference between atomic mass and gram atomic mass?

Atomic mass is a pure number for one atom (relative to one-twelfth of carbon-12). Gram atomic mass is that number in grams and is the mass of one mole of atoms.

Q: Does 22.4 L apply to all substances?

No. It is the volume of one mole of a gas at S.T.P. only. Liquids and solids have much smaller molar volumes.

Q: Why do equal volumes of hydrogen and carbon dioxide have different masses?

By Avogadro's law they contain the same number of molecules, but each CO₂ molecule (44) is 22 times heavier than an H₂ molecule (2).

Q: Do I need to prove molar volume or atomicity for ICSE?

No. The syllabus says formal proofs are not required; you should be able to state the laws, explain atomicity using equations such as the formation of HCl, and do simple calculations.

Updated 15 October 2026 · Fact-checked

A mole is the amount of substance containing 6.022 × 10²³ particles (Avogadro's number). One mole of a substance has a mass equal to its gram atomic or gram molecular mass, and one mole of any gas occupies 22.4 litres at S.T.P., its molar volume. These three facts link mass, number and volume.

Notice that one mole links all three: its mass in grams (18 g for water), 6.022 × 10²³ particles, and 22.4 L if it is a gas at STP.

Illustration drawn for QuestionBuddy.

Understand Mole, Avogadro Number and Molar Volume

Chemists cannot count atoms one by one, so they count them in packs, just as eggs are counted in dozens. The pack is the mole: $6.022 \times 1 0^{23}$ particles, a number called Avogadro's number ( $N_{A}$ ). The particles may be atoms, molecules or ions, so always say which: 1 mole of oxygen atoms is $6.022 \times 1 0^{23}$ O atoms (16 g), but 1 mole of oxygen molecules is $6.022 \times 1 0^{23}$ O₂ molecules (32 g).

Mass of one mole. The atomic mass of an element is a number for one atom, relative to $\frac{1}{12}$ of a carbon-12 atom. The same number in grams is the gram atomic mass, the mass of 1 mole of atoms (1 gram atom). Likewise the molecular mass is a number for one molecule, and the gram molecular mass is the mass of 1 mole of molecules (1 gram molecule). For water, molecular mass = 18, gram molecular mass = 18 g.

Avogadro's law. Equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules. It follows that 1 mole of every gas takes up the same volume at a given temperature and pressure.

Molar volume. The volume occupied by 1 mole (1 gram molecular mass) of any gas at S.T.P. (0 °C or 273 K, and 1 atm or 760 mm of mercury) is 22.4 litres. Put another way, the mass of 22.4 L of any gas at S.T.P. equals its gram molecular mass: 22.4 L of hydrogen weighs 2 g, of oxygen 32 g, of carbon dioxide 44 g. Molar volume applies only to gases: a mole of water at room temperature is about 18 cm³ of liquid, not 22.4 L.

Atomicity from volumes. Avogadro's law explains why hydrogen, oxygen, nitrogen and chlorine are diatomic. Experiment shows that 1 volume of hydrogen and 1 volume of chlorine give 2 volumes of hydrogen chloride. If 1 volume holds $n$ molecules, $n$ molecules of hydrogen give $2 n$ molecules of HCl, each containing at least one hydrogen atom. So each hydrogen molecule must split into 2 atoms: hydrogen is H₂. The same reasoning with ammonia ( $N_{2} + 3 H_{2} \to 2 N H_{3}$ ) and nitric oxide ( $N_{2} + O_{2} \to 2 N O$ ) shows N₂ and O₂ are diatomic. (The syllabus asks for this explanation, not a formal proof.)

The triangle of conversions. Moles sit at the centre: number $= n \times N_{A}$ , mass $= n \times M$ , gas volume at S.T.P. $= n \times 22.4$ L. Move between any two by going through moles.

Mole, Avogadro Number and Molar Volume: formulas and key facts

Moles from mass: $n = \frac{mass (g)}{gram molecular (or atomic) mass}$
Moles from number of particles: $n = \frac{N}{6.022 \times 1 0 ^{23}}$
Moles of a gas at S.T.P.: $n = \frac{V ( L )}{22.4 L} = \frac{V ( cm ^{3} )}{22 400 cm ^{3}}$
Number of atoms in a molecular sample: $atoms = n \times N_{A} \times atomicity$; Atomicity: H₂, O₂, N₂, Cl₂ = 2; O₃ = 3; NH₃ = 4 atoms per molecule.
One mole of a gas at S.T.P.: Hydrogen H₂ 2 g in 22.4 L
Oxygen O₂ 32 g in 22.4 L
Nitrogen N₂ 28 g in 22.4 L
Chlorine Cl₂ 71 g in 22.4 L
Carbon dioxide CO₂ 44 g in 22.4 L
Ammonia NH₃ 17 g in 22.4 L
Standard temperature and pressure (S.T.P.): Temperature 0 °C or 273 K
Pressure 1 atm or 760 mm of mercury
Molar volume 22.4 litres per mole of gas

Hydrogen H₂	2 g in 22.4 L
Oxygen O₂	32 g in 22.4 L
Nitrogen N₂	28 g in 22.4 L
Chlorine Cl₂	71 g in 22.4 L
Carbon dioxide CO₂	44 g in 22.4 L
Ammonia NH₃	17 g in 22.4 L

Temperature	0 °C or 273 K
Pressure	1 atm or 760 mm of mercury
Molar volume	22.4 litres per mole of gas

How to solve Mole, Avogadro Number and Molar Volume questions

Mole numericals are conversions. Put moles in the middle and every question becomes two short steps.

1Identify what is given (mass, number of particles or gas volume at S.T.P.) and what is asked.
2Convert the given quantity to moles using the matching formula.
3Convert moles to the asked quantity: × gram molecular mass for grams, × 6.022 × 10²³ for particles, × 22.4 L for a gas at S.T.P.
4If atoms are asked for a molecular substance, multiply the number of molecules by the atomicity.
5To compare samples, convert each to moles (or moles of atoms) and compare those numbers.
6Write units at every line: mol, g, L or cm³, molecules or atoms.

Mole, Avogadro Number and Molar Volume in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Volume to mass and number for a gas

Suppose a party balloon is filled with 4.48 L of helium, measured at S.T.P. Helium exists as single atoms (He = 4).

What mass of helium is in the balloon, and how many helium atoms does it hold?

Try it first, then show the solutionSolution

Moles $= 4.48/22.4 = 0.2$ mol.
Mass $= 0.2 \times 4 = 0.8$ g.
Atoms $= 0.2 \times 6.022 \times 1 0^{23} = 1.2044 \times 1 0^{23}$ .

Answer: 0.8 g of helium, about 1.20 × 10²³ atoms.

Memory hook: A balloon of gas is just moles in disguise: divide the litres by 22.4.

Mass to volume: how much space a gas needs

Suppose a hospital oxygen cylinder holds 0.5 kg of oxygen gas compressed. Assume the gas were released and measured at S.T.P. (O = 16).

What volume would this oxygen occupy at S.T.P.?

Try it first, then show the solutionSolution

Moles of O₂ $= 500/32 = 15.625$ mol.
Volume $= 15.625 \times 22.4 = 350$ L.

Answer: 350 L at S.T.P., which is why gases are stored compressed.

Memory hook: 32 grams of oxygen fills 22.4 litres; 500 g fills about 16 such boxes.

Tricks to remember Mole, Avogadro Number and Molar Volume

Short memory aids. Say each one aloud once, then use it on a question the same day.

Moles in the middle
Draw a triangle with 'mol' at the centre and mass, number and volume at the corners. Multiply going out (× M, × N_A, × 22.4); divide coming in.
Use it when: Every mole conversion.
Same T, same P, same N
Avogadro's law: same volume + same temperature + same pressure = same number of molecules.
Use it when: Writing the full statement without missing the conditions.
Molar volume needs three words
One mole, gas, S.T.P.: a definition of molar volume without all three loses the mark.
Use it when: Defining molar volume.

Mole, Avogadro Number and Molar Volume: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Which sample has the most molecules? · 1 mark · MCQBoard level

Which of the following contains the largest number of molecules? (H = 1, C = 12, N = 14, O = 16)
(a) 2 g of hydrogen
(b) 22 g of carbon dioxide
(c) 33.6 L of nitrogen at S.T.P.
(d) 17 g of ammonia

Try it first, then show the solutionSolution

Moles: (a) $2/2 = 1$ ; (b) $22/44 = 0.5$ ; (c) $33.6/22.4 = 1.5$ ; (d) $17/17 = 1$ .
Most moles means most molecules.

Answer: (c) 33.6 L of nitrogen (1.5 mol).

Example 2: Molecules and atoms in ammonia · 2 marks · Short answerBoard level

Calculate the number of molecules and the total number of atoms in 3.4 g of ammonia. (N = 14, H = 1, $N_{A} = 6.022 \times 1 0^{23}$ )

Try it first, then show the solutionSolution

Moles of NH₃ $= 3.4/17 = 0.2$ mol.
Molecules $= 0.2 \times 6.022 \times 1 0^{23} = 1.2044 \times 1 0^{23}$ .
Each molecule has 4 atoms: atoms $= 4 \times 1.2044 \times 1 0^{23} = 4.8176 \times 1 0^{23}$ .

Answer: 1.20 × 10²³ molecules; 4.82 × 10²³ atoms.

Example 3: Mass and molecules from a gas volume · 3 marks · Short answerHarder board level

Find the mass and the number of molecules in 5.6 L of sulphur dioxide at S.T.P. (S = 32, O = 16)

Try it first, then show the solutionSolution

Moles $= 5.6/22.4 = 0.25$ mol.
Gram molecular mass of SO₂ $= 32 + 32 = 64$ g; mass $= 0.25 \times 64 = 16$ g.
Molecules $= 0.25 \times 6.022 \times 1 0^{23} = 1.5055 \times 1 0^{23}$ .

Answer: 16 g; 1.51 × 10²³ molecules.

Example 4: Explaining that hydrogen is diatomic · 3 marks · Short answerHarder board level

One volume of hydrogen combines with one volume of chlorine to give two volumes of hydrogen chloride, all at the same temperature and pressure. Use Avogadro's law to explain why a hydrogen molecule must contain two atoms.

Try it first, then show the solutionSolution

By Avogadro's law, if 1 volume contains $n$ molecules, then $n$ molecules of H₂ and $n$ of Cl₂ give $2 n$ molecules of HCl.
Each HCl molecule contains at least one hydrogen atom, so $2 n$ HCl molecules contain at least $2 n$ hydrogen atoms.
These came from only $n$ hydrogen molecules, so each hydrogen molecule contains 2 atoms: $H_{2} + C l_{2} \to 2 H C l$ .

Answer: Hydrogen is diatomic (H₂): $n$ molecules supply $2 n$ atoms to $2 n$ HCl molecules.

Example 5: Equal masses, different gases · 3 marks · Short answerHOTS / competency

Equal masses of oxygen (O₂) and sulphur dioxide (SO₂) are taken at S.T.P. Find the ratio of (a) their volumes, (b) the numbers of atoms they contain. (S = 32, O = 16)

Try it first, then show the solutionSolution

Let the mass be $m$ g. Moles: O₂ $= m /32$ ; SO₂ $= m /64$ .
(a) Volume ratio = mole ratio $= \frac{m /32}{m /64} = 2 : 1$ .
(b) Atoms: O₂ has 2 per molecule, SO₂ has 3. Ratio $= \frac{2 \times m /32}{3 \times m /64} = \frac{4}{3}$ , so 4 : 3.

Answer: (a) 2 : 1 (b) 4 : 3.

Example 6: Identifying a gas from its mass and volume · 4 marks · Long answerHOTS / competency

1.42 g of a gaseous element occupies 448 cm³ at S.T.P. (a) Find its gram molecular mass and vapour density. (b) Which of O₂, Cl₂ or SO₂ could it be? (c) How many atoms does the sample contain? (O = 16, S = 32, Cl = 35.5)

Try it first, then show the solutionSolution

Moles $= 448/22 400 = 0.02$ mol.
(a) Gram molecular mass $= 1.42/0.02 = 71$ g; vapour density $= 71/2 = 35.5$ .
(b) O₂ = 32, Cl₂ = 71, SO₂ = 64 (and SO₂ is a compound, not an element): the gas is chlorine.
(c) Cl₂ is diatomic: atoms $= 0.02 \times 2 \times 6.022 \times 1 0^{23} = 2.4088 \times 1 0^{22}$ .

Answer: (a) 71 g; VD 35.5 (b) Chlorine, Cl₂ (c) 2.41 × 10²² atoms.

Where students lose marks in Mole, Avogadro Number and Molar Volume

The slips examiners see most often, and what to write instead.

Mistake: Incomplete definition of molar volume
In 2024 (Q4(i)) examiners noted that candidates left out 'S.T.P.' and 'gas', or confused molar volume with relative molecular mass.
Wrong working
Molar volume is the volume of one mole of a substance.
Right working
Molar volume is the volume occupied by one mole of any gas at S.T.P., i.e. 22.4 L.
Do this: Include all three: one mole, gas, S.T.P.
Usually costs: 1 mark.
Mistake: Avogadro's law without conditions
In 2025 (Q8(i)) examiners reported that candidates omitted 'under the same conditions of temperature and pressure'.
Do this: Equal volumes of all gases, under the same temperature and pressure, contain the same number of molecules.
Usually costs: 1 mark.
Mistake: Guessing in mole MCQs
In 2024 (Q1(xiii)) examiners noted that options were guessed without calculating moles for each item.
Do this: Convert every option to moles in the margin; it takes under a minute.
Usually costs: 1 mark.
Mistake: Atoms versus molecules
Students give the number of molecules when atoms are asked.
Wrong working
Atoms in 0.2 mol NH₃ = 1.2044 × 10²³
Right working
Atoms = 4 × 1.2044 × 10²³ = 4.8176 × 10²³
Do this: Multiply by the atomicity when atoms are asked.
Usually costs: 1 mark.

Exam tips

Learn the definitions of mole, gram atom, gram molecule, molar volume and Avogadro's law word for word; 'state the term' questions carry 5 marks in Section A.
Use $N_{A} = 6.022 \times 1 0^{23}$ (or the value printed in the paper).
22.4 L is for gases at S.T.P. only; never use it for water, solids or a gas at room conditions unless told.
Write answers in standard form with three or four significant figures.

Practice questions on Mole, Avogadro Number and Molar Volume

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The volume occupied by 44 g of carbon dioxide at S.T.P. is:
(a) 11.2 L
(b) 22.4 L
(c) 44.8 L
(d) 4.48 L
Hint
44 g is one gram molecular mass of CO₂.
Answer
(b) 22.4 L.
Q2 · 1 mark · Very short answerBoard level
Define molar volume.
Hint
Three key words: one mole, gas, S.T.P.
Answer
The volume occupied by one mole (one gram molecular mass) of any gas at S.T.P.; it is 22.4 litres.
Q3 · 2 marks · Short answerBoard level
How many moles and how many atoms are present in 4.6 g of sodium? (Na = 23, $N_{A} = 6.022 \times 1 0^{23}$ )
Hint
Sodium is counted as atoms.
Answer
Moles $= 4.6/23 = 0.2$ mol; atoms $= 0.2 \times 6.022 \times 1 0^{23} = 1.2044 \times 1 0^{23}$ .
Q4 · 2 marks · Short answerHarder board level
Calculate the mass of $3.011 \times 1 0^{22}$ molecules of water. (H = 1, O = 16)
Hint
Number → moles → mass.
Answer
Moles $= \frac{3.011 \times 1 0 ^{22}}{6.022 \times 1 0 ^{23}} = 0.05$ mol; mass $= 0.05 \times 18 = 0.9$ g.
Q5 · 2 marks · Short answerHarder board level
1.12 L of a gas at S.T.P. has a mass of 2.2 g. Find its gram molecular mass and its vapour density.
Hint
Mass of 22.4 L = gram molecular mass.
Answer
Mass of 22.4 L $= 2.2 \times \frac{22.4}{1.12} = 44$ g, so gram molecular mass = 44 g; vapour density $= 44/2 = 22$ .
Q6 · 1 mark · Assertion-reasonHarder board level
Assertion (A): 1 mole of oxygen atoms and 1 mole of oxygen molecules have the same mass.
Reason (R): Both contain $6.022 \times 1 0^{23}$ particles.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What is the mass of each?
Answer
(d). R is true, but 1 mole of O atoms is 16 g while 1 mole of O₂ molecules is 32 g.
Q7 · 3 marks · Short answerHarder board level
Arrange in increasing order of the number of molecules: 8 g of methane; 4.48 L of oxygen at S.T.P.; 1.8 g of water; $1.8066 \times 1 0^{23}$ molecules of nitrogen. (C = 12, H = 1, O = 16)
Hint
Convert all four to moles.
Answer
Moles: CH₄ 0.5; O₂ 0.2; H₂O 0.1; N₂ 0.3. Order: water < oxygen < nitrogen < methane.
Q8 · 3 marks · Short answerHOTS / competency
Which contains more atoms, 14 g of nitrogen gas or 8.5 g of ammonia? Show working. (N = 14, H = 1)
Hint
Count moles of atoms, not molecules.
Answer
N₂: 0.5 mol molecules × 2 = 1 mol of atoms. NH₃: 0.5 mol molecules × 4 = 2 mol of atoms ( $1.2044 \times 1 0^{24}$ ). Ammonia contains twice as many atoms, although both samples have the same number of molecules.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Value of Avogadro's number?Show answer
6.022 × 10²³ per mole.
2.Mass of 22.4 L of ammonia at S.T.P.?Show answer
17 g (its gram molecular mass).
3.Number of molecules in 0.5 mol of CO₂?Show answer
3.011 × 10²³.
4.What is a gram atom?Show answer
One mole of atoms of an element; its mass is the gram atomic mass.
5.Moles in 11.2 L of any gas at S.T.P.?Show answer
0.5 mol.
6.State the values of S.T.P.Show answer
0 °C (273 K) and 1 atm (760 mm of mercury).

Mole, Avogadro Number and Molar Volume: frequently asked questions

What is the difference between atomic mass and gram atomic mass?

Atomic mass is a pure number for one atom (relative to one-twelfth of carbon-12). Gram atomic mass is that number in grams and is the mass of one mole of atoms.

Does 22.4 L apply to all substances?

No. It is the volume of one mole of a gas at S.T.P. only. Liquids and solids have much smaller molar volumes.

Why do equal volumes of hydrogen and carbon dioxide have different masses?

By Avogadro's law they contain the same number of molecules, but each CO₂ molecule (44) is 22 times heavier than an H₂ molecule (2).

Do I need to prove molar volume or atomicity for ICSE?

No. The syllabus says formal proofs are not required; you should be able to state the laws, explain atomicity using equations such as the formation of HCl, and do simple calculations.