The Mole and Avogadro Constant Exam Questions
Practice IGCSE Chemistry exam questions on the mole and Avogadro constant. Covers mole calculations, converting between mass and moles, molar gas volume, and the Avogadro number with mark schemes and examiner notes.
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Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
These questions cover all aspects of the mole and Avogadro constant. Show all working in calculations.
Question 1 (2 marks, Supplement)
Define the term “mole” as used in chemistry.
Mark scheme
- The amount of substance that contains 6.02 x 10^23 particles [1]
- (Avogadro’s constant / number) / the same number of particles as there are atoms in 12 g of carbon-12 [1]
Examiner note: “A mole is a lot of atoms” scores 0. The Avogadro constant (6.02 x 10^23) or the carbon-12 reference must appear in your answer.
Question 2 (3 marks, Supplement)
Calculate the number of moles of each substance. (Ar: H = 1, O = 16, Na = 23, S = 32)
(a) 46 g of sodium (Na) (1)
(b) 4.0 g of hydrogen gas (H2) (1)
(c) 16 g of sulfur dioxide (SO2) (1)
Mark scheme
(a) n = mass / Ar = 46 / 23 = 2.0 mol [1]
(b) n = mass / Mr = 4.0 / (2 x 1) = 4.0 / 2 = 2.0 mol [1]
(c) n = mass / Mr = 16 / (32 + 32) = 16 / 64 = 0.25 mol [1]
Examiner note: For elements that exist as diatomic molecules (H2, O2, N2, Cl2, etc.), use Mr not Ar. Hydrogen gas is H2 with Mr = 2, not H with Ar = 1. This is a very common error.
Question 3 (3 marks, Supplement)
Calculate the mass of:
(a) 0.5 mol of calcium carbonate, CaCO3 (2)
(b) 0.1 mol of sulfuric acid, H2SO4 (1)
(Ar: H = 1, C = 12, O = 16, S = 32, Ca = 40)
Mark scheme
(a) Mr of CaCO3 = 40 + 12 + (3 x 16) = 100 [1] mass = n x Mr = 0.5 x 100 = 50 g [1]
(b) Mr of H2SO4 = (2 x 1) + 32 + (4 x 16) = 98 mass = 0.1 x 98 = 9.8 g [1]
Examiner note: Rearrange n = mass/Mr to mass = n x Mr. Always calculate Mr first, then multiply by the number of moles. Include units (g) in your final answer.
Question 4 (3 marks, Supplement)
At room temperature and pressure (RTP), the molar gas volume is 24 dm3.
(a) Calculate the volume of 2 mol of carbon dioxide gas at RTP. (1)
(b) Calculate the number of moles in 6.0 dm3 of oxygen gas at RTP. (1)
(c) Calculate the volume occupied by 3.2 g of methane (CH4) at RTP. (1)
(Ar: H = 1, C = 12)
Mark scheme
(a) V = n x 24 = 2 x 24 = 48 dm3 [1]
(b) n = V / 24 = 6.0 / 24 = 0.25 mol [1]
(c) Mr of CH4 = 12 + 4 = 16; n = 3.2 / 16 = 0.2 mol; V = 0.2 x 24 = 4.8 dm3 [1]
Examiner note: The molar gas volume (24 dm3 at RTP or 24,000 cm3) applies to ALL gases regardless of their identity. One mole of any gas occupies 24 dm3 at RTP. For (c), convert mass to moles first, then moles to volume.
Question 5 (3 marks, Supplement)
Calculate the number of molecules in 9.0 g of water (H2O). (Avogadro constant = 6.02 x 10^23, Ar: H = 1, O = 16)
Mark scheme
- Mr of H2O = 18 [1]
- n = 9.0 / 18 = 0.5 mol [1]
- Number of molecules = 0.5 x 6.02 x 10^23 = 3.01 x 10^23 [1]
Examiner note: The sequence is: mass → moles (using Mr) → number of particles (using Avogadro constant). To go from moles to particles, multiply by 6.02 x 10^23. To go from particles to moles, divide.
Question 6 (4 marks, Supplement)
A sample of gas has a volume of 480 cm3 at RTP.
(a) Convert 480 cm3 to dm3. (1)
(b) Calculate the number of moles of gas. (1)
(c) If the gas is carbon dioxide (CO2), calculate its mass. (2)
(Ar: C = 12, O = 16)
Mark scheme
(a) 480 / 1000 = 0.48 dm3 [1]
(b) n = 0.48 / 24 = 0.02 mol [1]
(c) Mr of CO2 = 12 + 32 = 44 [1] mass = 0.02 x 44 = 0.88 g [1]
Examiner note: Unit conversion is tested frequently: 1 dm3 = 1000 cm3. If the volume is given in cm3, convert to dm3 before using the molar gas volume (24 dm3), or use the molar gas volume in cm3 (24,000 cm3). Be consistent.
Question 7 (2 marks, Supplement)
Explain why 2 g of hydrogen gas (H2) and 32 g of oxygen gas (O2) contain the same number of molecules.
(Ar: H = 1, O = 16)
Mark scheme
- 2 g of H2 = 2/2 = 1 mol; 32 g of O2 = 32/32 = 1 mol [1]
- Both contain 1 mole, so both contain 6.02 x 10^23 molecules / the same number of molecules [1]
Examiner note: One mole of any substance contains the same number of particles (Avogadro’s number). The masses are different because the Mr values are different, but the number of moles — and therefore the number of molecules — is the same.
Question 8 (4 marks, Supplement)
A student has 4.48 dm3 of an unknown gas at RTP. The mass of the gas is 8.8 g.
(a) Calculate the number of moles of gas. (1)
(b) Calculate the molar mass (Mr) of the gas. (1)
(c) The gas contains carbon and oxygen only. Deduce the molecular formula of the gas. (2)
(Ar: C = 12, O = 16)
Mark scheme
(a) n = 4.48 / 24 = 0.187 mol (accept 0.19) [1]
(b) Mr = mass / n = 8.8 / 0.187 = 47.1 (accept rounding to allow 44 if n = 0.2) [1]
(c) If Mr = 44: C + O = 12 + 32 = 44, giving CO2… but 12 + 32 = 44 works for CO2 [1] Molecular formula = CO2 [1]
Examiner note: This is a reverse calculation — using volume and mass to find Mr, then using Mr to deduce the formula. The gas is carbon dioxide. If your Mr is slightly off due to rounding, check whether rounding moles more carefully gives 44 (4.48/24 = 0.1867, and 8.8/0.1867 = 47.1 — accept answers leading to CO2).
What to revise if you scored below 6
If the mole calculations were difficult, memorise the triangle: n = mass / Mr. For gas volume problems, remember: volume = n x 24 dm3 at RTP. Practise converting between cm3 and dm3. Revisit the mole and Avogadro constant notes.
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