Year 11 Chemistry Module 2 ⏱ ~35 min Lesson 6 of 20 IQ2 Begins

Concentration

A cup of coffee and an espresso both contain caffeine — but one hits harder. That difference is concentration. Every time a nurse measures an IV drip rate, a pool technician tests chlorine levels, or a chemist prepares a reagent, they're working with this concept. It is the single most used calculation in Year 11 and 12 Chemistry.

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📐

Formula Reference — This Lesson

c = n ÷ V

c = concentration (mol L⁻¹) n = moles of solute (mol) V = volume of solution (L)

Find c: c = n ÷ V | Find n: n = c × V | Find V: V = n ÷ c

⚠️ Critical habit: V must be in litres (L) before substituting. If given mL, divide by 1000 first. This is the #1 source of errors in concentration calculations.

📖 Know

Key Facts

Definition: concentration = moles of solute ÷ volume of solution
Units: mol L⁻¹ (also written M or mol/L)
Solute, solvent, solution — key terms

💡 Understand

Concepts

Why concentration depends on both amount AND volume
Why mL must be converted to L before calculating
Difference between mol L⁻¹ and g L⁻¹

✅ Can Do

Skills

Calculate c, n, or V using c = n ÷ V
Convert g L⁻¹ to mol L⁻¹ and back
Solve multi-step problems involving concentration + molar mass

📚 Core Content

🔬

What Is Concentration?

A solution is a homogeneous mixture of a solute (the substance being dissolved) in a solvent (the dissolving medium, usually water). Concentration tells you how much solute is packed into a given volume of solution.

In chemistry, concentration is almost always expressed in moles per litre (mol L⁻¹) — also called molarity. This unit connects the macroscopic world (litres of solution you can measure in a lab) to the microscopic world (moles of particles) via a single formula.

Dilute

0.1 mol L⁻¹

Few solute particles
per litre of solution

Moderate

0.5 mol L⁻¹

Moderate number
of solute particles

Concentrated

2.0 mol L⁻¹

Many solute particles
per litre of solution

The Critical Unit Conversion

The volume in c = n ÷ V must be in litres. Laboratory volumes are almost always given in millilitres. This conversion must become automatic.

250 mL

millilitres

→

÷ 1000

0.250 L

litres ← use this

Why this matters: If you substitute 250 mL directly instead of 0.250 L, your concentration will be 1000 times too small. A student who gets the formula right but forgets the conversion will lose marks every time. Build the habit: see mL → divide by 1000 → then substitute.

Concentration in g L⁻¹ vs mol L⁻¹

Sometimes concentration is given or required in grams per litre (g L⁻¹). To convert between the two, use molar mass as the bridge.

g L⁻¹ → mol L⁻¹: divide by MM
c (mol L⁻¹) = c (g L⁻¹) ÷ MM

mol L⁻¹ → g L⁻¹: multiply by MM
c (g L⁻¹) = c (mol L⁻¹) × MM

🧮 Worked Examples

Worked Example 1 — Finding concentration

Stepwise

0.500 mol of sodium hydroxide (NaOH) is dissolved to make 250 mL of solution. Calculate the concentration in mol L⁻¹.

1

Identify and convert units

n = 0.500 mol | V = 250 mL

V = 250 ÷ 1000 = 0.250 L

⚠️ Convert mL → L before substituting
2

Apply formula

c = n ÷ V = 0.500 ÷ 0.250

c = 2.00 mol L⁻¹

✓ Answerc = 2.00 mol L⁻¹

Worked Example 2 — Finding volume

Stepwise

What volume of 2.00 mol L⁻¹ hydrochloric acid (HCl) contains 0.300 mol of HCl?

1

Identify known quantities

c = 2.00 mol L⁻¹ | n = 0.300 mol | V = ?
2

Rearrange and solve

V = n ÷ c = 0.300 ÷ 2.00 = 0.150 L

= 150 mL

✓ AnswerV = 0.150 L (150 mL)

Worked Example 3 — Multi-step: mass → moles → concentration

Stepwise

5.85 g of sodium chloride (NaCl) is dissolved in water to make 500 mL of solution. Calculate the concentration in mol L⁻¹. (Na = 22.990, Cl = 35.453)

1

Calculate molar mass

MM(NaCl) = 22.990 + 35.453 = 58.443 g mol⁻¹
2

Convert mass → moles

n = m ÷ MM = 5.85 ÷ 58.443 = 0.1001 mol
3

Convert volume and find concentration

V = 500 ÷ 1000 = 0.500 L

c = n ÷ V = 0.1001 ÷ 0.500 = 0.200 mol L⁻¹

✓ Answerc = 0.200 mol L⁻¹

⚠️

Common Mistakes

Substituting mL instead of L

c = 0.5 ÷ 250 = 0.002 mol L⁻¹ ✗ — answer is 1000 times too small. The unit mol L⁻¹ demands litres in the denominator.

✓ Fix: Always write the unit conversion step explicitly: 250 mL ÷ 1000 = 0.250 L, then substitute 0.250.

Confusing the volume of solute with the volume of solution

If you dissolve 50 mL of ethanol in 200 mL of water, the solution volume is NOT 200 mL — it's approximately 250 mL (though not exactly, due to volume contraction). The formula uses the volume of the total solution, not the solvent alone.

✓ Fix: The V in c = n/V is the volume of the final solution — the number you read on the volumetric flask or measuring cylinder after everything is mixed and made up to volume.

Using g L⁻¹ concentration directly in c = n/V

If a question gives concentration in g L⁻¹, you cannot substitute it directly — c = n/V only works with mol L⁻¹. You must first convert g L⁻¹ to mol L⁻¹ by dividing by MM.

✓ Fix: c (mol L⁻¹) = c (g L⁻¹) ÷ MM. Always check units before substituting.

📓 Copy Into Your Books

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📖 Definitions

Solute — substance being dissolved
Solvent — dissolving medium (usually water)
Solution — homogeneous mixture of solute in solvent
Concentration (c) — moles of solute per litre of solution

📐 Formula & Rearrangements

c = n ÷ V (find concentration)
n = c × V (find moles)
V = n ÷ c (find volume)
Units: mol L⁻¹ | mol | L

⚠️ Critical Habit

V must be in litres — always convert mL ÷ 1000
c (mol L⁻¹) = c (g L⁻¹) ÷ MM
V in formula = volume of solution, not solvent

🔗 Multi-Step Pathway

Mass → moles: n = m ÷ MM
Moles + volume → concentration: c = n ÷ V
Concentration + volume → moles: n = c × V

🧪 Activities

📊 Activity 1 — Practice Drill

Applying c = n ÷ V

Four problems — all three rearrangements plus a unit conversion problem. Show working before revealing answers.

1 Calculate the concentration of a solution made by dissolving 0.400 mol of glucose in 2.00 L of solution.

n = 0.400 mol, V = 2.00 L (already in litres) c = n ÷ V = 0.400 ÷ 2.00 = 0.200 mol L⁻¹
2 How many moles of KNO₃ are in 400 mL of a 0.250 mol L⁻¹ solution?

Convert: V = 400 mL ÷ 1000 = 0.400 L n = c × V = 0.250 × 0.400 = 0.100 mol
3 What volume of 0.500 mol L⁻¹ sulfuric acid contains 0.0750 mol of H₂SO₄?

c = 0.500 mol L⁻¹, n = 0.0750 mol V = n ÷ c = 0.0750 ÷ 0.500 = 0.150 L = 150 mL
4 A solution of copper sulfate (CuSO₄) has a concentration of 16.0 g L⁻¹. Express this in mol L⁻¹. (Cu = 63.546, S = 32.06, O = 15.999)

MM(CuSO₄) = 63.546 + 32.06 + 4(15.999) = 159.602 g mol⁻¹ c (mol L⁻¹) = 16.0 ÷ 159.602 = 0.1002 mol L⁻¹

Type your working for all four problems:

Answer in your workbook.

✏️ Complete in your workbook

🔢 Activity 2 — Data Analysis

Complete the Concentration Table

The table below shows solution data. Calculate the missing values (shown as ?). Show all working in the space below.

Solution	Moles of solute (mol)	Volume (mL)	Volume (L)	Concentration (mol L⁻¹)
NaOH	0.500	250	?	?
HCl	?	100	?	2.00
CaCl₂	0.0400	?	?	0.160
H₂SO₄	1.50	600	?	?
KMnO₄	?	50.0	?	0.0200

Show all working below:

Complete in your workbook — show all working.

✏️ Complete table in your workbook

❓ Multiple Choice

🎯

Test Your Knowledge

1. Which expression correctly calculates concentration in mol L⁻¹ when given 0.30 mol of solute in 150 mL of solution?

c = 0.30 ÷ 150 = 0.0020 mol L⁻¹

c = 0.30 × 150 = 45.0 mol L⁻¹

c = 0.30 ÷ 0.150 = 2.00 mol L⁻¹

c = 150 ÷ 0.30 = 500 mol L⁻¹

2. A 500 mL solution contains 0.250 mol of KCl. What is its concentration?

0.125 mol L⁻¹

0.500 mol L⁻¹

1.00 mol L⁻¹

2.00 mol L⁻¹

3. How many moles of NaOH are in 25.0 mL of a 0.100 mol L⁻¹ solution?

2.50 × 10⁻³ mol

0.250 mol

2.50 mol

4.00 × 10⁻³ mol

4. A student dissolves 4.00 g of NaOH in water and makes 200 mL of solution. What is the concentration? (Na = 22.990, O = 15.999, H = 1.008)

0.0500 mol L⁻¹

0.100 mol L⁻¹

0.400 mol L⁻¹

0.500 mol L⁻¹

5. A solution of glucose (C₆H₁₂O₆) has a concentration of 18.0 g L⁻¹. What is this in mol L⁻¹? (MM of glucose = 180.16 g mol⁻¹)

10.0 mol L⁻¹

0.180 mol L⁻¹

0.100 mol L⁻¹

1.80 mol L⁻¹

✍️ Short Answer

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Extended Questions

6. A nurse prepares an intravenous saline solution by dissolving 9.00 g of NaCl in water to produce 1.00 L of solution. Calculate the concentration of the solution in (a) mol L⁻¹ and (b) g L⁻¹. (Na = 22.990, Cl = 35.453) 4 MARKS

Type your answer below:

Answer in your workbook.

✏️ Answer in your workbook

7. A chemist needs to deliver 0.0500 mol of H₂SO₄ to a reaction. They have a stock solution of 1.00 mol L⁻¹ sulfuric acid. What volume of stock solution should they measure out? Give your answer in mL. 3 MARKS

Type your answer below:

Answer in your workbook.

✏️ Answer in your workbook

8. A pool technician tests a swimming pool and finds that the chlorine concentration is 3.55 g L⁻¹ of Cl₂ (dissolved chlorine gas). The pool holds 50 000 L of water. Calculate the total mass of Cl₂ in the pool, and the number of Cl₂ molecules present. (Cl = 35.453) 5 MARKS

Type your answer below:

Answer in your workbook.

✏️ Answer in your workbook

✅ Comprehensive Answers

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📊 Activity 2 — Table Answers

NaOH: V = 250÷1000 = 0.250 L; c = 0.500÷0.250 = 2.00 mol L⁻¹

HCl: V = 100÷1000 = 0.100 L; n = c×V = 2.00×0.100 = 0.200 mol

CaCl₂: V = n÷c = 0.0400÷0.160 = 0.250 L = 250 mL

H₂SO₄: V = 600÷1000 = 0.600 L; c = 1.50÷0.600 = 2.50 mol L⁻¹

KMnO₄: V = 50.0÷1000 = 0.0500 L; n = c×V = 0.0200×0.0500 = 1.00×10⁻³ mol

❓ Multiple Choice

1. C — Must convert 150 mL → 0.150 L first. c = 0.30 ÷ 0.150 = 2.00 mol L⁻¹.

2. B — V = 0.500 L. c = 0.250 ÷ 0.500 = 0.500 mol L⁻¹.

3. A — V = 0.0250 L. n = 0.100 × 0.0250 = 2.50 × 10⁻³ mol.

4. D — MM(NaOH) = 39.997 g mol⁻¹. n = 4.00 ÷ 39.997 = 0.100 mol. V = 0.200 L. c = 0.100 ÷ 0.200 = 0.500 mol L⁻¹.

5. C — c = 18.0 ÷ 180.16 = 0.100 mol L⁻¹.

📝 Short Answer Model Answers

Q6 (4 marks):

MM(NaCl) = 22.990 + 35.453 = 58.443 g mol⁻¹ n = m ÷ MM = 9.00 ÷ 58.443 = 0.1540 mol (a) c = n ÷ V = 0.1540 ÷ 1.00 = 0.154 mol L⁻¹ (b) c (g L⁻¹) = 9.00 g ÷ 1.00 L = 9.00 g L⁻¹ (or: 0.154 × 58.443 = 9.00 g L⁻¹)

Q7 (3 marks):

V = n ÷ c = 0.0500 ÷ 1.00 = 0.0500 L = 50.0 mL

Q8 (5 marks):

Total mass = c × V = 3.55 × 50 000 = 177 500 g = 177.5 kg MM(Cl₂) = 2 × 35.453 = 70.906 g mol⁻¹ n = m ÷ MM = 177 500 ÷ 70.906 = 2503 mol N = n × Nₐ = 2503 × 6.022 × 10²³ = 1.507 × 10²⁷ molecules

Mark lesson as complete

Tick when you've finished all activities and checked your answers.

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