Module 2 · L6 of 20~35 min⚡ +50 XP in Learn · +25 to completeIQ3 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.
Today's hook, A teaspoon of sugar in a cup, or the same teaspoon in a swimming pool, the amount is the same, but the experience couldn't be more different. What's the maths behind that gap?
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You're here
Warm up and think first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
If you dissolve one teaspoon of sugar in a small cup of water versus the same teaspoon in a swimming pool, the amount of sugar is the same, but something is very different. What is different, and how would you describe that difference using a mathematical expression? What two quantities would you need to measure?
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The formula, and the words for it
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Formula reference · this lesson
core formula
$$c = \dfrac{n}{V}$$
c = concentration (mol L⁻¹)
n = moles of solute (mol)
V = volume of solution (L)
Find c: $c = n \div V$
Find n: $n = c \times V$
Find V: $V = n \div c$
⚠️ 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.
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What you'll master
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 $c = n \div V$
Convert g L⁻¹ to mol L⁻¹ and back
Solve multi-step problems involving concentration + molar mass
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Key terms
Concentration (c)
Amount of solute per unit volume of solution; standard unit is mol L⁻¹ (molarity).
Solute
The substance that is dissolved in a solution (e.g. NaCl in saltwater).
Solvent
The dissolving medium, usually water in aqueous solutions.
c = n ÷ V
Concentration formula: c (mol L⁻¹) = n (mol) ÷ V (L). Volume must be in litres.
g L⁻¹
Mass concentration; convert to mol L⁻¹ by dividing by molar mass: c (mol L⁻¹) = c (g L⁻¹) ÷ M.
Aqueous solution
A solution in which water is the solvent; denoted (aq) after the formula.
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What concentration actually measures
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What is concentration?
core concept
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.
The critical unit conversion
The volume in $c = n \div V$ must be in litres. Laboratory volumes are almost always given in millilitres. This conversion must become automatic.
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
Concentration: c = n ÷ V (mol L⁻¹), where n is in mol and V must be in litres (divide mL by 1000). Solute = dissolved substance; solvent = dissolving medium. To interconvert g L⁻¹ and mol L⁻¹, divide or multiply by molar mass.
Pause, copy the highlighted definition into your book before moving on.
Did you get this? True or false: 250 mL of a 0.40 mol L⁻¹ solution contains 0.10 mol of solute.
Quick check: A solution made by dissolving 0.100 mol of NaCl in 200 mL of solution has what concentration?
Two truths, one lie about concentration in mol L⁻¹. Pick the lie.
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Concentration Explorer
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Interactive · Concentration Explorer
try it
Adjust moles and volume to see how concentration changes in real time.
↗ Drag to rotate · Scroll to zoom · Double-click to resetInteractive 3D
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Three worked examples
Worked examples · reveal as you go
Worked example 1 · finding concentration+5 XP on full reveal
0.500 mol of sodium hydroxide (NaOH) is dissolved to make 250 mL of solution. Calculate the concentration in mol L⁻¹.
Convert mass to moles: n = m ÷ MM = 5.85 ÷ 58.443 = 0.1001 mol.
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The three traps that cost marks
Common errors · the 3 traps that cost marks
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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.
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Confusing volume of solute with volume of solution
If you dissolve 50 mL of ethanol in 200 mL of water, the solution volume is NOT 200 mL, it's about 250 mL. The formula uses the volume of the total solution, not the solvent alone.
Fix: V is the volume of the final solution, the number you read on the volumetric flask after everything is mixed and made up to volume.
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Using g L⁻¹ 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 by dividing by MM.
Fix: c (mol L⁻¹) = c (g L⁻¹) ÷ MM. Always check units before substituting.
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Quick-fire drill, then revisit your thinking
Quick-fire practice · 5 reps +2 XP per reveal
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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. c = n ÷ V = 0.400 ÷ 2.00 = 0.200 mol L⁻¹
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How many moles of KNO₃ are in 400 mL of a 0.250 mol L⁻¹ solution?
V = 400 ÷ 1000 = 0.400 L. n = c × V = 0.250 × 0.400 = 0.100 mol
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What volume of 0.500 mol L⁻¹ sulfuric acid contains 0.0750 mol of H₂SO₄?
V = n ÷ c = 0.0750 ÷ 0.500 = 0.150 L (150 mL)
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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 = 16.0 ÷ 159.602 = 0.1002 mol L⁻¹
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A nurse prepares 250 mL of saline by dissolving 2.25 g of NaCl. What is the concentration in mol L⁻¹? (Na = 22.990, Cl = 35.453)
MM(NaCl) = 58.443. n = 2.25 ÷ 58.443 = 0.0385 mol. V = 0.250 L. c = 0.0385 ÷ 0.250 = 0.154 mol L⁻¹
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Revisit your thinking
At the start of this lesson, you described the difference between dissolving sugar in a small cup versus a swimming pool, and thought about what quantities you'd need to measure.
The answer: concentration = amount of solute ÷ volume of solution, or $c = n \div V$ (in mol L⁻¹). The amount of sugar (solute) stays the same, but the volume of the solution is much larger in the pool, so the concentration is far lower. You need to measure both the number of moles of solute and the volume in litres.
Use the Stoichiometry Calculator. How many moles are in 44 g of CO₂ (molar mass = 44 g/mol)?
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Practice questions
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Multiple choice
+2 XP per correct · +5 bonus if perfect
Pick your answer, then rate your confidence. That tells the system what to drill next.
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Short answer
ApplyLower-order4 marks
Q1. 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)
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ApplyLower-order3 marks
Q2. 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.
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AnalyseMid-order5 marks
Q3. A concentrated chlorine dosing stock contains 3.55 g L⁻¹ of Cl₂. A 50 000 L batch of this stock is prepared. Calculate the total mass of Cl₂ in the batch and the number of Cl₂ molecules present. (Cl = 35.453) Note: 3.55 g L⁻¹ is an illustrative dosing-stock figure for the calculation; pool water that swimmers contact is kept at only about 1–3 mg L⁻¹ of free available chlorine (the active disinfectant), which is not the same quantity as total Cl₂.
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EvaluateHigher-order5 marks
Q4. A student prepares a solution by dissolving 5.85 g of NaCl in 100 mL of water. They claim the concentration is 1.00 mol L⁻¹. A second student says: "You made the solution incorrectly, you should have added water to the NaCl to produce exactly 100 mL of solution, not dissolved it in 100 mL of water." Evaluate both the concentration claim and the procedural claim. (Na = 22.990, Cl = 35.453)
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CreateHigher-order7 marks
Q5. A scientist needs to prepare 500 mL of a 0.500 mol L⁻¹ solution of potassium permanganate (KMnO₄) as a stock solution. They then plan to transfer two 50.0 mL aliquots, one for a titration and one to compare colour intensity. (a) Calculate the mass of KMnO₄ needed. (b) Outline a complete procedure to prepare the stock solution. (c) Calculate how many moles of KMnO₄ are in each 50.0 mL aliquot. (K = 39.098, Mn = 54.938, O = 15.999)
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📖 Comprehensive answers (click to reveal)
Multiple choice
Convert millilitres to litres first. Must convert 150 mL → 0.150 L first. c = 0.30 ÷ 0.150 = 2.00 mol L⁻¹.
Concentration from moles and volume. V = 0.500 L. c = 0.250 ÷ 0.500 = 0.500 mol L⁻¹.
Moles from concentration and volume. V = 0.0250 L. n = 0.100 × 0.0250 = 2.50 × 10⁻³ mol.
Mass to moles to concentration. 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.
Grams per litre to moles per litre. c = 18.0 ÷ 180.16 = 0.100 mol L⁻¹.
Short answer model answers
Q1 (4 marks): MM(NaCl) = 22.990 + 35.453 = 58.443 g mol⁻¹ [1]. n = 9.00 ÷ 58.443 = 0.1540 mol [1]. (a) c = 0.1540 ÷ 1.00 = 0.154 mol L⁻¹ [1]. (b) c (g L⁻¹) = 9.00 g ÷ 1.00 L = 9.00 g L⁻¹ [1].
Q2 (3 marks): V = n ÷ c = 0.0500 ÷ 1.00 = 0.0500 L = 50.0 mL.
Q3 (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 = 177 500 ÷ 70.906 = 2503 mol. N = n × Nₐ = 2503 × 6.022 × 10²³ = 1.507 × 10²⁷ molecules.
Q4 (5 marks):Concentration check: MM(NaCl) = 58.443; n = 5.85 ÷ 58.443 = 0.1001 mol [1]. If total V ≈ 100 mL: c = 0.1001 ÷ 0.100 = 1.00 mol L⁻¹ [1], only true if adding NaCl to 100 mL of water doesn't change the total volume [1]. Procedural evaluation: The second student is correct [1]. For accurate prep, dissolve NaCl in a small volume, transfer to a 100 mL volumetric flask, and make up to the mark [1].
Q5 (7 marks): (a) MM(KMnO₄) = 39.098 + 54.938 + 4(15.999) = 158.032 g mol⁻¹ [1]. n = c × V = 0.500 × 0.500 = 0.250 mol [1]. m = 0.250 × 158.032 = 39.51 g [1]. (b) Weigh 39.51 g of KMnO₄. Dissolve in ~300 mL distilled water in a beaker with stirring. Quantitatively transfer to a 500 mL volumetric flask, rinsing 3× with small volumes. Make up to the 500 mL mark, stopper and invert [2]. (c) n = c × V = 0.500 × 0.0500 = 0.0250 mol per aliquot [2]. Note: KMnO₄ is not a primary standard, it is difficult to obtain perfectly pure (it usually contains some MnO₂) and slowly decomposes in light, so its exact concentration is uncertain from weighing alone. For accurate work the prepared solution is standardised by titration against a true primary standard (e.g. sodium oxalate) before use.
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Retrieve, then finish
Check what actually stuck
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
