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hscscience Chem · Y11
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Module 2 · L20 of 20 ~30 min ⚡ +15 XP in Learn · +25 to complete

Module 2 Review

The complete reference for Module 2, Introduction to Quantitative Chemistry. Every formula, every key term, every exam question type, and worked examples of each. Use this as your study reference before the module quiz and the HSC.

Today's hook, The complete reference for Module 2, Introduction to Quantitative Chemistry. Every formula, every key term, every exam question type, and worked examples of each. Use this as your study reference before the module quiz and the HSC.
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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

01
Recall, your gut answer first
+5 XP warm-up

Without looking at any notes, can you list the four key formulas that connect moles (n) to mass, particles, gas volume, and concentration? And which single concept acts as the "universal bridge" that links every calculation in this module?

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03
What you'll master
Know

Key facts

  • The four Module 2 inquiry questions and what each covers
  • Core formulas: n = m ÷ MM, N = n × Nₐ, V = n × Vₘ, n = c × V, PV = nRT
  • STP = 22.71 L/mol; SATP = 24.8 L/mol; concordant titres within 0.10 mL
Understand

Concepts

  • How the mole links mass, particles, gas volume and concentration
  • When to apply purity, percentage yield, limiting reagent and the gas laws
  • Why a true yield is ≤ 100% and an apparent value above 100% signals error
Can do

Skills

  • Solve multi-step problems that chain several inquiry questions
  • Choose the correct molar volume and convert temperatures to kelvin
  • Check answers for physical plausibility (yield ≤ 100%, concordant titres)
04
Key terms
Mole (mol)
SI unit of amount of substance; 1 mol = 6.022 × 10²³ elementary entities (Avogadro's constant).
Molar mass (M)
Mass per mole of a substance in g mol⁻¹; numerically equal to the relative atomic or molecular mass.
Concentration (c)
Amount of solute per unit volume; c = n/V in mol L⁻¹; ppm and ppb used for trace concentrations.
Stoichiometric ratio
The ratio of moles of substances from the balanced equation; used to convert between moles of different species.
Limiting reagent
Determines the maximum theoretical yield; identified by the smallest mole-to-coefficient ratio.
Gravimetric and volumetric analysis
Two quantitative techniques: gravimetric uses mass of precipitate; volumetric uses volume of titrant at the equivalence point.
1

Misconception to fix

Wrong: Concentration and amount of solute are the same thing.

2

Misconception to fix

Right: Concentration is amount per unit volume; the same amount of solute can produce different concentrations in different volumes.

3

Applying purity and yield at the wrong stage

Percentage purity must be applied to the impure sample mass before converting to moles, while percentage yield is applied only at the very end (to the theoretical product). Students often calculate moles from the full impure mass, or multiply by yield too early, which makes every downstream value wrong.

Fix: Follow the order: impure mass → pure mass (× % purity) → moles → mole ratio → theoretical product → × % yield. Purity first, yield last.

Work mode · how are you completing this lesson?
1

Calculate the mass of 0.25 mol of calcium carbonate, CaCO₃. (M = 100.09)

2

A solution contains 0.40 mol of NaCl dissolved in 250 mL. Calculate its concentration in mol L⁻¹.

3

A 3.0 L sample of gas at 100 kPa is compressed at constant temperature until the pressure is 150 kPa. Find the new volume.

4

How many molecules are present in 8.0 g of methane, CH₄? (M = 16.0)

5

Complete combustion CH₄ + 2O₂ → CO₂ + 2H₂O: what volume of CO₂ (25 °C, 100 kPa) forms from 0.50 mol of CH₄? (Vₘ = 24.79 L mol⁻¹)

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12
Revisit your thinking

At the start of this lesson, you tried to recall the four key formulas and identify the concept that bridges all of Module 2.

The mole (n) is the universal bridge, every formula in Module 2 converts something into moles or from moles: n = m ÷ MM (mass), N = n × Nₐ (particles), V = n × molar volume (gas), and n = c × V (solution). Every calculation, whether IQ1, IQ2, IQ3 or IQ4 (the gas laws, where n appears in PV = nRT), passes through n as the central intermediate step. If you can automatically and accurately perform each of these four conversions, the rest of every stoichiometry or concentration problem follows.

Reflect: how did your initial thinking compare to what you've learned?

Write a reflection in your workbook.

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Interactive Tool, Stoichiometry Calculator Open fullscreen ↗
Use the Stoichiometry Calculator. How many moles are in 44 g of CO₂ (molar mass = 44 g/mol)?
01
Multiple choice
+2 XP per correct · +5 bonus if perfect

Pick your answer, then rate your confidencethat tells the system what to drill next.

02
Short answer
01
Boss battle
earn bronze · silver · gold

Five timed questions on module 2 review. Beat the boss to bank a tier, gold (perfect + fast), silver (80%+), or bronze (cleared).

⚔ Enter the arena
02
Science Jump · Module 2 Review
arcade practice

Climb platforms, hit checkpoints, and answer questions on this lesson's topic.

Mark lesson as complete

Tick when you've finished the practice and review.