Chemistry · Year 11 · Module 2 · Lesson 1
HSC Exam Practice
The Mole Concept
Short answer
1.Short answer
Define mole and Avogadro’s constant. In your answer, state the units of Avogadro’s constant and explain why those units are appropriate.
Distinguish between N (capital) and n (lowercase) in the formula N = n × NA. In your answer, state the units of each quantity and give a numerical example that illustrates the difference.
Explain why chemists use the mole as a unit for measuring amounts of substance rather than counting individual atoms. In your answer, refer to the scale of atoms and to Avogadro’s number.
Calculate the number of atoms in 4.5 mol of sodium (Na). Show all working, including the formula used, values substituted, and the final answer with appropriate units or absence of units.
A student argues that “1 mol of carbon atoms and 1 mol of oxygen molecules (O2) are the same because they both contain one mole of substance, so they must contain identical numbers of atoms.” Outline why this argument is partly correct and partly incorrect. Identify the number of atoms in 1 mol of O2 as part of your answer.
Describe how the definition of the mole changed in 2019. In your answer, state what the new definition is based on, how this differs from the previous definition (which referenced carbon-12), and whether this change affects practical Year 11 mole calculations.
Data response
2.Data response — moles and particle counts across five laboratory samples
A chemistry technician prepares five samples for a school experiment and records the amount of each substance in moles. The table below shows the data collected.
| Sample | Substance | n (mol) | Elementary entity | N (number of entities) |
|---|---|---|---|---|
| P | Neon atoms (Ne) | 1.50 | atom | |
| Q | Water molecules (H2O) | molecule | 1.806 × 1024 | |
| R | Sodium chloride (NaCl) formula units | 0.750 | formula unit | |
| S | Helium atoms (He) | atom | 3.011 × 1023 | |
| T | Carbon atoms (C) | 4.00 | atom |
NA = 6.022 × 1023 mol−1
(a) Complete the missing cells in the table. Show one full worked example of your calculation. (5 marks: 1 per missing cell; 1 worked example)
(b) Identify the sample with the greatest number of entities and the sample with the fewest. Justify your choices by comparing their values of N. (2 marks)
The graph below shows the number of entities (N) plotted against the amount in moles (n) for five samples of different substances. Each data point represents a different laboratory sample.
(a) Describe the relationship between N and n shown in the graph. (1 mark)
(b) Identify the gradient of the line of best fit and state what physical quantity it represents. Give the units. (2 marks)
Extended response
3.Extended response
Analyse the role of the mole concept and Avogadro’s number in connecting the atomic scale to the laboratory scale. In your response, evaluate how the choice of 6.022 × 1023 as the value of NA — rather than any other number — makes the mole system particularly useful for chemists. Discuss the significance of the N vs n distinction for avoiding errors in stoichiometric calculations, and assess what would happen to mole calculations if the convention of writing capital N for particle count and lowercase n for amount were reversed or abandoned.
Chemistry · Year 11 · Module 2 · Lesson 1
Answer Key & Marking Guidelines
Section 1 · Short answer · 3 marks · Band 3
Sample response. The mole is the SI base unit for amount of substance; symbol n. One mole contains exactly 6.022 × 1023 elementary entities (atoms, molecules, ions, or formula units). Avogadro’s constant (NA) is 6.022 × 1023 mol−1. The units mol−1 are appropriate because NA gives the number of entities per mole — when you multiply n (in mol) by NA (in mol−1), the mol units cancel and you obtain a dimensionless count N.
Marking notes. 1 mark for correctly defining the mole (SI unit of amount of substance, contains NA entities); 1 mark for correctly defining Avogadro’s constant (6.022 × 1023 mol−1); 1 mark for explaining why the units mol−1 are appropriate (they cancel with mol to give a dimensionless count).
Section 1 · Short answer · 4 marks · Band 3–4
Sample response. N (capital) is the actual number of individual particles in a sample — a pure, dimensionless count with no units (e.g. N = 1.204 × 1024 atoms). n (lowercase) is the amount of substance in moles, with units of mol (e.g. n = 2 mol). They are linked by N = n × NA. Example: for 2 mol of carbon, N = 2 × 6.022 × 1023 = 1.204 × 1024 atoms (capital N, no units); n = 2 mol (lowercase, units mol). Confusing them is the single most common error in mole calculations.
Marking notes. 1 mark for correctly describing N (particle count, no units); 1 mark for correctly describing n (amount in mol); 1 mark for correct numerical example showing both values for the same sample; 1 mark for explicitly linking them via N = n × NA.
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. Individual atoms are so small that they cannot be counted directly or weighed individually — a single carbon atom has a mass of approximately 2 × 10−23 g, far below the sensitivity of any laboratory balance. The mole groups 6.022 × 1023 particles together into an amount large enough to weigh on a lab balance. For example, one mole of carbon atoms has a mass of exactly 12 g — a measurable quantity. Avogadro’s number (NA) is the bridge between the atomic scale (individual particles) and the laboratory scale (grams and millilitres).
Marking notes. 1 mark for explaining that atoms are too small to count individually or to weigh directly (with quantitative reference e.g. ~10−23 g per atom); 1 mark for stating that the mole groups NA particles into a measurable amount; 1 mark for explicitly naming Avogadro’s number as the bridge between atomic and laboratory scales (or linking it to molar mass = 12 g/mol for carbon-12).
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. Known: n = 4.5 mol, NA = 6.022 × 1023 mol−1. Formula: N = n × NA. Substitute: N = 4.5 × 6.022 × 1023. Calculate: N = 2.710 × 1024 atoms. Note: N has no units (the mol units cancel: mol × mol−1 = dimensionless).
Marking notes. 1 mark for correctly writing the formula N = n × NA; 1 mark for correct substitution (4.5 × 6.022 × 1023); 1 mark for correct final answer (2.71 × 1024 atoms, no units — accept 2.7 × 1024). Deduct 1 mark if units are attached to N.
Section 1 · Short answer · 3 marks · Band 4
Sample response. The student is partly correct: both 1 mol C and 1 mol O2 contain exactly the same number of molecules/elementary entities (NA = 6.022 × 1023). However, the statement that they contain identical numbers of atoms is incorrect. Each O2 molecule contains 2 oxygen atoms, so 1 mol of O2 molecules contains 2 × NA = 2 × 6.022 × 1023 = 1.204 × 1024 oxygen atoms. One mole of C contains NA = 6.022 × 1023 carbon atoms. Therefore 1 mol O2 contains twice as many atoms as 1 mol C.
Marking notes. 1 mark for correctly identifying the valid part (both contain NA elementary entities / molecules); 1 mark for explaining the error (O2 is diatomic so 1 mol O2 = 2 NA atoms); 1 mark for stating the correct number of O atoms in 1 mol O2 (1.204 × 1024 or 2 NA).
Section 1 · Short answer · 2 marks · Band 4
Sample response. Before 2019, the mole was defined as the amount of substance containing the same number of entities as atoms in exactly 12 g of carbon-12, which tied NA to a physical material. Since 2019, the mole is defined by fixing NA exactly at 6.02214076 × 1023 mol−1 as a counting number, removing dependence on any physical artefact. This change has no practical effect on Year 11 calculations: the value of NA changed by less than 1 part in 108, which is far below the precision of any 3–4 significant figure calculation at Year 11 level.
Marking notes. 1 mark for correctly describing old definition (linked to carbon-12 physical sample) vs new definition (NA fixed as exact number); 1 mark for stating the new definition has no practical effect on Year 11 calculations (value changed <1 part in 108).
Section 2 · Data response · 7 marks · Band 4–5
Sample response (a) — missing cells.
P: N = 1.50 × 6.022 × 1023 = 9.033 × 1023 atoms. [1 mark]
Q: n = N ÷ NA = 1.806 × 1024 ÷ 6.022 × 1023 = 3.00 mol. [1 mark]
R: N = 0.750 × 6.022 × 1023 = 4.517 × 1023 formula units. [1 mark]
S: n = N ÷ NA = 3.011 × 1023 ÷ 6.022 × 1023 = 0.500 mol. [1 mark]
T: N = 4.00 × 6.022 × 1023 = 2.409 × 1024 atoms. [1 mark]
Worked example (Sample P): Known: n = 1.50 mol, NA = 6.022 × 1023 mol−1. Formula: N = n × NA. Substitute: N = 1.50 × 6.022 × 1023 = 9.033 × 1023 atoms. [The worked example mark is built into the 5 above; accept any sample shown clearly.]
Sample response (b). Greatest: Sample T (n = 4.00 mol) with N = 2.409 × 1024 entities [1]. Fewest: Sample S (n = 0.500 mol) with N = 3.011 × 1023 entities [1]. Because N = n × NA and NA is a fixed constant, N is directly proportional to n; the sample with the most moles always has the most entities regardless of which substance it is.
Marking notes. 1 mark per correct missing cell (P, Q, R, S, T — formula and answer must match the correct operation); 1 mark for identifying greatest (T) and fewest (S) with values as justification; 1 mark for comparison reasoning.
Section 2 · Data response · 3 marks · Band 4–5
Sample response (a). N and n show a directly proportional (linear) relationship passing through the origin: as n increases, N increases in direct proportion.
Sample response (b). The gradient = rise / run = ΔN / Δn. Reading from the graph, at n = 1 mol, N ≈ 6.022 × 1023. Therefore gradient ≈ 6.022 × 1023 mol−1. This is Avogadro’s constant, NA. Units: mol−1 (or entities per mole).
Marking notes. Part (a): 1 mark for directly proportional / linear through origin. Part (b): 1 mark for correctly identifying the gradient as NA = 6.022 × 1023 mol−1; 1 mark for correctly naming it as Avogadro’s constant with units mol−1.
Section 3 · Extended response · 7 marks · Band 5–6
Sample response. The mole concept exists to bridge the inaccessible atomic scale with the measurable laboratory scale. Individual atoms are on the order of 10−10 m in diameter and have masses of ~10−23 g — entirely beyond direct counting or weighing. By defining one mole as containing exactly NA = 6.022 × 1023 entities, chemists can work with measurable gram-scale quantities while knowing precisely how many particles are involved. The choice of 6.022 × 1023 specifically was not arbitrary: it was defined (originally) so that one mole of carbon-12 has a mass of exactly 12 g. This makes the molar mass (in g/mol) numerically equal to the relative atomic mass from the periodic table, so a chemist can instantly convert: relative atomic mass = molar mass in grams, no further arithmetic required. Any other value of NA would break this elegant equivalence. The N vs n distinction is essential for avoiding the most common stoichiometric errors. N is the actual particle count — a dimensionless integer on the order of 1023–1024; n is the amount in mol — a small number typically between 10−3 and 10. They differ by a factor of NA ≈ 6 × 1023, so confusing them leads to errors of 23–24 orders of magnitude. For example, a student who substitutes n = 2 into a formula expecting N will calculate 2 particles instead of 1.2 × 1024 particles — an error that makes all subsequent stoichiometry impossible. If the convention of uppercase/lowercase were abandoned, chemists would need to specify in words “number of particles” or “amount in moles” every time — increasing the probability of misinterpretation in equations and making dimensional analysis harder. The current convention is efficient precisely because it encodes the distinction into a single letter case: if units are present (mol), it is n; if there are no units, it is N. Abandoning this would increase cognitive load and error rates in multi-step calculations such as stoichiometry, dilution, and gas law problems. In summary, both the specific numerical value of NA and the N/n notation convention are deliberate, deeply practical choices that make the mole system work efficiently at the interface of atomic theory and laboratory chemistry.
Marking criteria (7 marks). 1 = explains why the mole bridges the atomic and laboratory scales (atoms too small; mole = measurable mass). 1 = correctly explains why NA = 6.022 × 1023 specifically (molar mass = relative atomic mass; carbon-12 = 12 g/mol). 1 = evaluates the usefulness of this specific value (makes conversion trivial; any other value would break the equivalence). 1 = correctly distinguishes N and n with reference to scale and units. 1 = explains the practical consequence of confusing N and n (order-of-magnitude error; example given). 1 = evaluates what abandoning the convention would mean (increased ambiguity, error, cognitive load — or a valid counterargument). 1 = overall synthesis: reaches an explicit judgement that both the value of NA and the notation convention are necessary features of a workable mole system, not arbitrary choices.