Using Concentrations of Solutions in mol:dm³ and Volumes of Gases [chemistry only, HT only]

AQA 8462 — 4.3.4 & 4.3.5

⬢ HIGHER TIER · CHEMISTRY ONLY ⬢

Concentrations in mol/dm³

& Volumes of Gases

The HT-only quantitative finale of Module 3. Two formulas: c = n ÷ V (for solutions) and V = n × 24 (for gases at RTP). Two recurring traps: forgetting the cm³ → dm³ conversion, and over-complicating gas-phase calculations when the volume ratio shortcut works directly.

What You'll Learn

Part 1 — Solutions in mol/dm³ (4.3.4)

Calculate concentration in mol/dm³ using c = n ÷ V. Find moles or mass of solute given concentration and volume. Convert between g/dm³ and mol/dm³. Use titration data to find an unknown concentration.

Part 2 — Volumes of Gases (4.3.5)

Apply Avogadro's principle: equal moles of gases occupy equal volumes at the same T and P. Use the molar gas volume = 24 dm³/mol at RTP to convert between mass, moles and gas volumes. Apply volume = mole ratio as a shortcut for gas-phase reactions.

Why It Matters

This lesson is calculation-heavy and worth roughly 6–10 marks per series. The single most-rewarded skill: writing every step on its own line so method marks are protected even if the final answer slips.

Molar Concentration

You met g/dm³ in Lesson 3.05 — that measures the mass of solute per dm³. The HT version uses moles instead, because moles connect directly to chemical equations and mole ratios.

Concentration in mol/dm³

The number of moles of solute dissolved per dm³ of solution.

concentration (mol/dm³) = moles of solute ÷ volume of solution (dm³)

Three Forms of the Formula

c = n ÷ V | n = c × V | V = n ÷ cVolume must always be in dm³. If a question gives you cm³, convert first: V (dm³) = V (cm³) ÷ 1000.

Worked Example 1

1

Calculate Molar Concentration

0.500 dm³ of solution contains 0.0750 mol of HCl. Calculate the concentration in mol/dm³. [2 marks]

Step 1 — Identify values: n = 0.0750 mol; V = 0.500 dm³ (already in dm³)

Step 2 — Apply formula: c = n ÷ V

Step 3 — Substitute: c = 0.0750 ÷ 0.500

c = 0.150 mol/dm³

Mark Scheme

2 marks: 1 for method (c = n ÷ V or substitution); 1 for correct value with unit. Always write the unit — "mol/dm³" or "mol dm⁻³" both accepted.

Converting g/dm³ ↔ mol/dm³

Both units measure the same physical thing — they differ only in whether you count solute by mass or by moles. Switch between them using M_r.

concentration (g/dm³) = concentration (mol/dm³) × M_r

Why It Works

1 mole of any substance has a mass of M_r grams. So if there are n mol per dm³, there are n × M_r grams per dm³.

g/dm³ → mol/dm³

Divide by M_r:

4.0 g/dm³ HCl ÷ 36.5 = 0.110 mol/dm³

mol/dm³ → g/dm³

Multiply by M_r:

0.10 mol/dm³ NaOH × 40 = 4.0 g/dm³

Pick the Right Unit

Most exam questions use mol/dm³ because it's directly compatible with mole ratios from balanced equations. Convert to g/dm³ only when the question asks, or when you need a mass of solute.

Worked Example 2 — Titration

Real exam context (P1 Jun24 Q09.4): you're given moles of one reactant and the concentration of the other; find the volume needed for a complete reaction.

2

Find Volume of Acid in a Titration

A student titrated 0.0045 mol of NaOH with 0.15 mol/dm³ hydrochloric acid:

NaOH + HCl → NaCl + H₂O

Calculate the volume of HCl needed, in cm³. [2 marks]

Step 1 — Mole ratio: NaOH : HCl = 1 : 1, so moles of HCl = 0.0045 mol

Step 2 — Rearrange formula: V = n ÷ c

Step 3 — Substitute: V = 0.0045 ÷ 0.15 = 0.030 dm³

Step 4 — Convert to cm³: 0.030 × 1000 = 30 cm³

Volume of HCl = 30 cm³

Examiner Alert — P1 Jun24 Q09.4 (verbatim)

"A common error was to multiply the moles and concentration together." — i.e. using V = n × c instead of V = n ÷ c. Always rearrange from c = n ÷ V carefully: if you want V, divide n by c.

The mark scheme accepts an alternative method: convert concentration to mol/cm³ (0.15 ÷ 1000 = 0.00015), then V = 0.0045 ÷ 0.00015 = 30 cm³. But the examiner report flags a recurring error: students "multiplied by 1000 in the first step instead of dividing", getting 150 mol/cm³. Stick to the safer method on the left — convert volume at the end, not concentration at the start.

Spec Variation — Find Unknown Concentration

Spec 4.3.4 also specifies: "If the volumes of two solutions that react completely are known and the concentration of one solution is known, the concentration of the other solution can be calculated." Same logic, different unknown:

Use known volume × known concentration to find moles of one reactant.

Apply the mole ratio from the balanced equation to find moles of the other reactant.

Divide moles by the other volume (in dm³) to find the unknown concentration.

Avogadro's Principle

Gases behave very differently from solids and liquids when it comes to amount. Two containers of completely different gases at the same temperature and pressure contain the same number of molecules if they have the same volume.

Avogadro's Principle (spec wording)

"Equal amounts in moles of gases occupy the same volume under the same conditions of temperature and pressure."

Why This Is Powerful

For solids and liquids, equal moles ≠ equal volumes (1 mol of lead is far smaller than 1 mol of water). For gases, the molecules are so far apart that the type of gas doesn't matter — only how many molecules. The volume is determined entirely by mole count, T, and P.

Equal Moles = Equal Volumes

Three different gases. Different molecule types. Same temperature, same pressure, same number of moles → same volume.

The Critical Insight

The masses are different (2 g, 32 g, 44 g) but the volumes are identical (24 dm³ each). For gases, count moles — never mass — when comparing volumes.

Molar Gas Volume at RTP

Molar Gas Volume

The volume of 1 mole of any gas at room temperature and pressure (RTP: 20°C and 1 atmosphere) is 24 dm³.

volume of gas (dm³) = moles of gas × 24

Three Forms of the Formula

V = n × 24 | n = V ÷ 24 | (rearrange as needed)Volume must be in dm³ when using 24 directly. If you have cm³, convert first OR use 24 000 cm³/mol.

When 24 Doesn't Apply

The 24 dm³/mol value is specific to RTP only (20°C, 1 atm). Industrial reactions often happen at very different temperatures and pressures, where 1 mole of gas occupies a very different volume. Use 24 only when the question states "at RTP" or gives 24 explicitly.

Worked Example 3 — Gas Volume from Mass

3

Combine mass → moles → volume

5.00 g of calcium carbonate fully decomposes. Calculate the volume of CO₂ produced at RTP.

CaCO₃ → CaO + CO₂. M_r(CaCO₃) = 100; molar gas volume = 24 dm³/mol. [3 marks]

Step 1 — Moles of CaCO₃: n = m ÷ M_r = 5.00 ÷ 100 = 0.0500 mol

Step 2 — Mole ratio: CaCO₃ : CO₂ = 1 : 1, so moles CO₂ = 0.0500 mol

Step 3 — Volume: V = n × 24 = 0.0500 × 24

Volume of CO₂ = 1.20 dm³ (= 1200 cm³)

The 3-Step Pattern

Almost every "mass → gas volume" question follows this pattern: (1) mass → moles using M_r, (2) apply mole ratio from balanced equation, (3) moles → volume using × 24. Label each step — three steps, three marks.

The Gas-Phase Shortcut

Avogadro's principle gives a powerful shortcut: for gases at the same T and P, the volume ratio equals the mole ratio. You don't need to calculate moles first — work directly with volumes.

For gases at the same T & P:

volume ratio = mole ratio (from balanced equation)

Example

For N₂(g) + 3H₂(g) → 2NH₃(g): the mole ratio is 1 : 3 : 2. So the volume ratio is also 1 : 3 : 2 — directly. If you start with 10 cm³ of N₂, you need 30 cm³ of H₂ and you make 20 cm³ of NH₃. No mole calculation needed.

Watch the State Symbols

The shortcut applies only to gases (g). Solids, liquids, and aqueous species don't follow Avogadro's principle. If a balanced equation includes a solid or liquid product (e.g. SiO₂(s) or H₂O(l)), exclude it from the gas-volume calculation.

Examiner Alert — P1 Jun22 Q07.6

A 4-mark gas-phase calculation that students consistently over-complicate. The simple ratio method scores all 4 marks; the long calculation scores the same 4 but loses marks to arithmetic slips.

"2Si₂H₆(g) + 7O₂(g) → 4SiO₂(s) + 6H₂O(g)

30 cm³ of Si₂H₆ reacts with 150 cm³ (an excess) of O₂. Calculate the total volume of gases present after the reaction. All gas volumes at the same T and P." 4 marks

The Volume-Ratio Method

O₂ used: mole ratio Si₂H₆ : O₂ = 2 : 7. So volume of O₂ reacting = 30 × (7/2) = 105 cm³.

O₂ excess (left over): 150 − 105 = 45 cm³.

H₂O(g) made: mole ratio Si₂H₆ : H₂O = 2 : 6 = 1 : 3. So volume = 30 × 3 = 90 cm³.

SiO₂ excluded — it's a solid (s), not a gas.

Total gas after = O₂ excess + H₂O(g) = 45 + 90 = 135 cm³.

Two Pitfalls in This Question

(1) Including SiO₂ in the gas total — it's flagged as (s) in the equation; gas volume only counts (g). (2) Trying to use 24 dm³/mol — the question states "all volumes at the same T and P" but does NOT say RTP. So 24 may not apply. The volume ratio shortcut works regardless of T and P.

Worked Example 4 — Volume → Moles

Real exam context (P2 Jun24 Q09.2): given a gas volume read off a graph, calculate moles. Reverse direction of WE3.

4

Calculate Moles from Gas Volume

14 cm³ of hydrogen gas is collected at room temperature and pressure. Calculate the number of moles of hydrogen. The volume of one mole of any gas at RTP is 24 dm³. [4 marks]

Step 1 — Convert volume: V = 14 ÷ 1000 = 0.014 dm³

Step 2 — Rearrange formula: n = V ÷ 24

Step 3 — Substitute: n = 0.014 ÷ 24

n = 5.8 × 10⁻⁴ mol (or 0.00058 mol)

Examiner Alert — P2 Jun24 Q09.2 (verbatim)

Examiner report: "A significant number of students did not recognise that the volume of the gas should be converted into dm³. A small number of students believed that 1 cm³ = 1000 dm³ resulting in an answer 10⁶ times too large."

The two errors to avoid:

No conversion at all — using 14 ÷ 24 directly = 0.58 mol (1000× too large).

Wrong-direction conversion — treating 1 cm³ as 1000 dm³ instead of 0.001 dm³ (10⁶× too large).

The mark scheme accepts an alternative: convert 24 dm³ to 24 000 cm³, then divide 14 cm³ by 24 000 — same answer, no unit conversion of the volume. Either method gets all 4 marks only if every step is shown.

Common Errors at a Glance

TopicLoses the markEarns the mark

Volume conversion (cm³ → dm³)c = 0.0045 ÷ 0.15 = 0.030 (left in dm³, no conversion).V = 0.030 dm³ × 1000 = 30 cm³.

Gas-phase ratioConverting to moles, multiplying by 24, then back to volumes — long, error-prone.Volume ratio = mole ratio directly. 30 × (7/2) = 105 cm³ in one line.

State symbolsIncluding SiO₂(s) or H₂O(l) in a gas-volume total.Only count species marked (g) in the equation.

Wrong formula direction"Volume = moles × concentration" — multiplying when you should divide.c = n ÷ V; rearrange carefully: V = n ÷ c, n = c × V.

Alternative-method trapConverting 0.15 mol/dm³ to mol/cm³ by ×1000 (gives 150 mol/cm³).Divide by 1000 to convert mol/dm³ → mol/cm³ (gives 0.00015 mol/cm³).

Using 24 outside RTPApplying 24 dm³/mol to industrial reactions or non-RTP conditions.Use 24 only when stated. Otherwise use volume ratio shortcut.

PEE Framework — Choosing a Method

Question: A student is given the equation N₂(g) + 3H₂(g) → 2NH₃(g) and asked: "What volume of H₂ reacts with 50 cm³ of N₂ at the same T and P?" Should they (a) convert volumes to moles using 24 dm³/mol then back to volumes, or (b) use the volume ratio directly?3 marks

P — PointThe student should use the volume ratio directly (method b).

E — EvidenceBy Avogadro's principle, gases at the same T and P have volume ratios equal to mole ratios. So the N₂ : H₂ ratio of 1 : 3 applies directly to volumes.

E — ExplanationVolume of H₂ = 50 × 3 = 150 cm³. Method (a) would give the same answer but with extra steps that risk arithmetic errors — and 24 dm³/mol may not even apply if the conditions aren't RTP.

When to Use Each Method

Use the volume-ratio shortcut when all relevant species are gases at the same T and P (no mass given). Use the full mass → moles → volume calculation when the question gives you a mass of a reactant or product.

Practice Question 1

250 cm³ of a sodium hydroxide solution contains 0.025 mol of NaOH. M_r(NaOH) = 40.

(a) Calculate the concentration in mol/dm³. 3 marks

(b) Convert this concentration to g/dm³. 2 marks

(c) 25.0 cm³ of this NaOH solution is exactly neutralised by 20.0 cm³ of dilute sulfuric acid:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Calculate the concentration of the sulfuric acid in mol/dm³. 3 marks

(a) M1

Convert volume: V = 250 ÷ 1000 = 0.25 dm³

(a) M2

Apply formula: c = n ÷ V = 0.025 ÷ 0.25

(a) M3

c = 0.10 mol/dm³

(b) M1

Apply conversion: c (g/dm³) = c (mol/dm³) × M_r = 0.10 × 40

(b) M2

c = 4.0 g/dm³

(c) M1

Moles of NaOH used: n = c × V = 0.10 × (25.0 ÷ 1000) = 0.0025 mol

(c) M2

Mole ratio NaOH : H₂SO₄ = 2 : 1, so moles of H₂SO₄ = 0.0025 ÷ 2 = 0.00125 mol

(c) M3

c(H₂SO₄) = n ÷ V = 0.00125 ÷ (20.0 ÷ 1000) = 0.00125 ÷ 0.020 = 0.0625 mol/dm³

Practice Question 2

0.48 g of magnesium reacts with excess hydrochloric acid:

Mg + 2HCl → MgCl₂ + H₂

A_r(Mg) = 24; molar gas volume at RTP = 24 dm³/mol.

Calculate the volume of H₂ gas produced at RTP, in cm³. 4 marks

M1

Calculate moles of Mg: n(Mg) = m ÷ A_r = 0.48 ÷ 24 = 0.020 mol

M2

Apply mole ratio: Mg : H₂ = 1 : 1, so n(H₂) = 0.020 mol

M3

Calculate volume: V = n × 24 = 0.020 × 24 = 0.48 dm³

M4

Convert to cm³: 0.48 × 1000 = 480 cm³

•

Final answer: 480 cm³

Summary

Concentration in mol/dm³ (4.3.4) HT

c = n ÷ V (in dm³). Always convert cm³ → dm³ first (÷ 1000). Switch between g/dm³ and mol/dm³ using g/dm³ = mol/dm³ × M_r. For titrations, use the mole ratio from the balanced equation, then rearrange to find volume or concentration.

Volumes of Gases (4.3.5) HT

Avogadro's principle: equal moles of gases occupy equal volumes at the same T and P. Molar gas volume at RTP = 24 dm³/mol. Use V = n × 24 to convert moles to volume. Gas-phase shortcut: for gases at the same T and P, the volume ratio equals the mole ratio — work directly with volumes.

Most Costly Errors

(1) Forgetting cm³ → dm³ conversion (1000× wrong). (2) Including solids/liquids in a gas-volume total. (3) Using 24 dm³/mol when conditions aren't RTP.

When to Use the Shortcut

Volume ratio shortcut works when all relevant species are gases at the same T and P. Skip it if the question gives you a mass — use mass → moles → volume instead.

Exam Tip

Label every step as its own line: "V = … dm³"; "n = … mol"; "V(gas) = n × 24 = …". This protects method marks even when arithmetic slips. Quote the unit on every numerical answer — mol/dm³, dm³, cm³, or mol depending on what's asked.