Chemistry • Year 11 • Module 2 • Lesson 6

Concentration: Moles Per Litre

Apply c = n ÷ V and its rearrangements to real data, real scenarios and a graph of concentration change — building the exam skills needed for Band 4–5 questions.

Apply · Data & Reasoning

1. Interpret a concentration data table — laboratory solutions

A student prepared five solutions in the laboratory. The table below records the moles of solute and the volume of each solution. Some cells are missing. 10 marks

Solution	Solute	Moles of solute (mol)	Volume (mL)	Concentration (mol L⁻¹)
A	NaOH	0.500	250
B	HCl	0.200	100
C	CaCl₂	0.0400		0.160
D	H₂SO₄	1.50	600
E	KMnO₄	1.00 × 10⁻³	50.0

1.1 Complete the “Volume (L)” and “Concentration (mol L⁻¹)” columns for solutions A, B, D and E. For solution C, calculate the missing volume. Show the conversion step for each row. 8 marks (2 per solution)

1.2 Identify which solution (A–E) has the highest concentration and explain in one sentence why comparing concentrations requires the volume to be in the same unit. 2 marks

Stuck? Convert each volume first (mL ÷ 1000), then apply c = n ÷ V or V = n ÷ c where needed.

2. Interpret a concentration–dilution graph

A laboratory preparation starts with a 2.00 mol L⁻¹ stock solution of sodium chloride (NaCl) and progressively adds water to dilute it. The graph below shows how the concentration changes as the total volume increases from 100 mL to 1000 mL (while keeping the moles of NaCl constant at 0.200 mol). 8 marks

Figure 2.1. Concentration of NaCl(aq) vs volume of solution (n = 0.200 mol NaCl constant throughout). Water is added progressively. Illustrative data.

2.1 Describe the shape of the graph and explain, using c = n ÷ V, why the concentration decreases at a rapidly decreasing rate as volume increases. 3 marks

2.2 Using the graph, estimate the concentration when the total volume is 400 mL. Verify your estimate by calculating c directly. 2 marks

2.3 A student claims: “Doubling the volume always halves the concentration.” Use values from the graph to evaluate whether this claim is always, sometimes or never true for this dilution. 3 marks

Stuck? For Q2.3 check: V=100 (c=2.00) doubled to V=200 (c=1.00) — does it halve? Try V=200 to V=400.

3. Compare mol L⁻¹ and g L⁻¹ across four features

Complete the two-column table. For each feature, write a concise description that contrasts the two concentration units. 8 marks (1 per cell)

Feature	mol L⁻¹ (molar concentration)	g L⁻¹ (mass concentration)
What it counts
Formula used
When it is most useful
How to convert to the other unit

Stuck? Revisit the “Concentration in g L⁻¹ vs mol L⁻¹” section and the Key Terms panel in the lesson.

4. Predict and justify — IV drip at Sydney Children’s Hospital

A nurse at Sydney Children’s Hospital needs to administer 0.0900 mol of NaCl to a patient over 2 hours using an IV saline solution. The ward stocks only two solutions: Solution X (0.150 mol L⁻¹ NaCl) and Solution Y (0.900 mol L⁻¹ NaCl). 7 marks

4.1 Calculate the volume of each solution that would deliver exactly 0.0900 mol of NaCl. Express your answers in mL. Show the formula used and all working. 4 marks

4.2 The nurse must deliver no more than 700 mL of fluid per dose. Predict which solution is appropriate and justify your prediction using your calculated volumes from 4.1. 2 marks

4.3 MM(NaCl) = 58.44 g mol⁻¹. What mass of NaCl is contained in the chosen volume from 4.2? 1 mark

Stuck? Use V = n ÷ c for each solution, then compare to the 700 mL limit.

Answers — Do not peek before attempting

Q1.1 — Concentration table

Solution A (NaOH): V = 250 ÷ 1000 = 0.250 L; c = 0.500 ÷ 0.250 = 2.00 mol L⁻¹.

Solution B (HCl): V = 100 ÷ 1000 = 0.100 L; c = 0.200 ÷ 0.100 = 2.00 mol L⁻¹.

Solution C (CaCl₂): V = n ÷ c = 0.0400 ÷ 0.160 = 0.250 L = 250 mL.

Solution D (H₂SO₄): V = 600 ÷ 1000 = 0.600 L; c = 1.50 ÷ 0.600 = 2.50 mol L⁻¹.

Solution E (KMnO₄): V = 50.0 ÷ 1000 = 0.0500 L; c = 1.00 × 10⁻³ ÷ 0.0500 = 0.0200 mol L⁻¹.

Marking: 1 mark for correct V(L) conversion + 1 mark for correct c per solution (or V for C).

Q1.2 — Highest concentration

Solution D has the highest concentration at 2.50 mol L⁻¹. Comparing concentrations requires volumes in the same unit (litres) because the unit mol L⁻¹ is defined per litre; mixing units would give meaningless comparisons.

Q2.1 — Shape of graph (3 marks)

The graph shows a curved, decreasing (reciprocal / hyperbolic) shape: concentration drops steeply at first (100–300 mL) and then more gradually as V increases. According to c = n ÷ V, with n constant, c is inversely proportional to V; equal increases in V have less effect at large V than at small V, hence the flattening curve [1 shape; 1 inverse proportionality; 1 explanation of why it flattens].

Q2.2 — Estimate and verify (2 marks)

From the graph at V = 400 mL, c ≈ 0.50 mol L⁻¹ [1 estimate]. Verification: V = 0.400 L; c = 0.200 ÷ 0.400 = 0.500 mol L⁻¹ [1 calculation].

Q2.3 — Evaluate the “doubling halves” claim (3 marks)

The claim is always true for this dilution, because c = n ÷ V is a direct reciprocal relationship: if V doubles, c halves [1 judgement]. Evidence: V = 100 mL, c = 2.00; V = 200 mL, c = 1.00 (halved) [1 example]. V = 200 mL, c = 1.00; V = 400 mL, c = 0.500 (halved) [1 second example confirming pattern]. This holds because n is constant in this experiment.

Q3 — Compare mol L⁻¹ and g L⁻¹

What it counts: mol L⁻¹ — moles (particles) per litre; g L⁻¹ — mass (grams) of solute per litre.

Formula used: mol L⁻¹ — c = n ÷ V; g L⁻¹ — c = m ÷ V (where m is mass in grams).

When most useful: mol L⁻¹ — chemical calculations involving stoichiometry, mole ratios, titrations; g L⁻¹ — clinical/industrial settings where mass is measured directly (e.g. saline bags labelled by mass per volume).

How to convert: mol L⁻¹ — multiply by MM to get g L⁻¹; g L⁻¹ — divide by MM to get mol L⁻¹.

Q4.1 — Volume of each solution (4 marks)

Solution X: V = n ÷ c = 0.0900 ÷ 0.150 = 0.600 L = 600 mL [2 marks: formula + correct answer].

Solution Y: V = n ÷ c = 0.0900 ÷ 0.900 = 0.100 L = 100 mL [2 marks: formula + correct answer].

Q4.2 — Which solution is appropriate? (2 marks)

Solution Y (100 mL) is appropriate [1]. Solution X requires 600 mL, which is within the 700 mL limit and would also be acceptable. However, Solution Y at 100 mL delivers the same dose in a much smaller volume, which is clinically safer (less fluid overload risk) and is well within the 700 mL limit [1 justification]. Accept Solution X if student justifies that 600 mL < 700 mL and is therefore within limit.

Q4.3 — Mass of NaCl (1 mark)

m = n × MM = 0.0900 × 58.44 = 5.26 g (same for either solution — both deliver 0.0900 mol). [Award 1 mark for correct calculation; accept any solution from 4.2.]