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S5 Chemistry: Molarity and Dilution — Tutor Lesson Plan (45 min, worked titration, four errors)
Students convert grams correctly and then lose the mark by leaving the volume in cm3. This lesson puts the unit conversion inside the formula so it cannot be forgotten.
Objectives — the student can convert between cm3 and dm3 without prompting, calculate concentration, moles or volume from c = n / V, use c1V1 = c2V2 for dilution, and carry a titration calculation from burette reading to the concentration of the unknown.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Moles | Moles in 4.0 g of NaOH, Mr 40? | 0.100 mol |
| Unit conversion | Convert 250 cm3 to dm3 | 0.250 dm3 |
| Rearranging | Make V the subject of c = n / V | V = n / c |
02 Lesson flow (45 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-5 min | Hook | Two beakers, same colour, different concentration, ask which is stronger | Vote before any formula |
| 5-13 min | Scene 1 | cm3 to dm3, and why 1000 cm3 is 1 dm3 | Convert five volumes aloud |
| 13-22 min | Scene 2 | c = n / V with 250 cm3 of 0.100 M | Get 0.025 mol and 1.00 g |
| 22-30 min | Scene 3 | Dilution: moles unchanged, volume changes | 0.100 M in 25 cm3 becomes 0.010 M |
| 30-39 min | Scene 4 | Titration: average the concordant readings, then use the mole ratio | 0.125 M acid |
| 39-45 min | Exit check | Three decisions in section 08 | Marked at once |
03 Teaching script (read it out)
Scene 1: The unit trap is the whole topic
Concentration in mol per dm3 needs the volume in dm3, so 250 cm3 is 0.250 dm3. Divide by 1000 once, at the moment the volume is written down, and never afterwards. A student who writes “250 cm3 = 250 dm3” has already lost the mark before any chemistry happens.
Drill: five conversions in thirty seconds: 25, 50, 100, 250, 500 cm3.
Scene 2: One triangle, three questions
c = n / V. With 250 cm3 of 0.100 M sodium hydroxide, n = c x V = 0.100 x 0.250 = 0.025 mol, which is 1.00 g of NaOH at Mr 40.0. The same triangle answers “what volume do I need” and “what is the concentration”, so no new formula is needed for each question type.
Ask: “If I add water, which of the three quantities stays the same?” Moles, because no solute is added.
Scene 3: Dilution and titration in one line each
Dilution: moles before equals moles after, so c1V1 = c2V2. Taking 25 cm3 of 0.100 M and making it up to 250 cm3 gives 0.100 x 0.025 / 0.250 = 0.010 M. Titration: 20 cm3 of acid neutralises 0.00250 mol of base in a 1:1 reaction, so c = 0.00250 / 0.0200 = 0.125 M. Use the average of concordant titres, not the first reading.
Warn: a 2:1 mole ratio is where most titration answers go wrong, so write the equation first.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Concentration | Find the concentration of 0.025 mol in 250 cm3. | Convert to dm3, then c = n / V | 0.100 M |
| Dilution | Dilute 25 cm3 of 0.100 M to 250 cm3. | c1V1 = c2V2 | 0.010 M |
| Titration | Find the acid concentration from titre and base moles. | Equation, mole ratio, then c = n / V | 0.125 M |
Sanity check (teach this): A concentration above 10 M or below 0.001 M in a school titration is almost always a volume left in cm3. Check the order: volume in dm3, moles, ratio, concentration. Also check that the average titre ignores the rough run.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Find the mass of NaOH needed for 500 cm3 of 0.200 M solution. | 0.100 mol, so 4.00 g | Volume converted to dm3 first |
| 20.0 cm3 of 0.150 M NaOH neutralises 25.0 cm3 of HCl. Find the acid concentration. | 0.00300 mol in 0.0250 dm3, so 0.120 M | Ratio 1:1 stated, then divided |
| What volume of water must be added to 50.0 cm3 of 0.400 M to make 0.100 M? | Dilute to 200 cm3, so add 150 cm3 | Final volume, not volume added, in the formula |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Leaving the volume in cm3 | Divide by 1000 to get dm3 | Write dm3 next to the number, not after the calculation |
| Using the volume of water added instead of the final volume | c1V1 = c2V2 uses the final total volume | Ask what the total volume is after mixing |
| Ignoring the mole ratio in a titration | Write the equation and read the ratio before dividing | Ask how many moles of acid react with one mole of base |
| Averaging the rough titre with the accurate ones | Average only the concordant titres | Ask which reading was rough and why it is excluded |
07 Homework (paper, section and question numbers)
- HKDSE Chemistry Paper 1 multiple-choice practice paper — Questions on solutions and volumetric analysis; then redo section 05 questions 1 to 3
- Secondary chemistry chapter index — Read the volumetric analysis chapter before the practical
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Convert 25.0 cm3 to dm3. | 0.0250 dm3 | All correct: move to titration curves |
| How many moles in 250 cm3 of 0.100 M? | 0.0250 mol | One wrong: repeat scene 1 |
| Why does dilution keep the moles constant? | No solute is added or removed, only solvent | Two wrong: repeat scene 3 |
09 Tutor FAQ
Should students use c1V1 = c2V2 everywhere?
Only for dilution or for a 1:1 reaction. For other ratios convert to moles first; the formula hides the ratio.
How many decimal places in a titration answer?
Match the burette: readings to 0.05 cm3, so three significant figures is the safe default.
Do they need to know primary standards?
Not for the calculation, but they should know why a standard solution is made up to the mark and mixed.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
