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S5 Chemistry | Molarity and Dilution prep guide (for tutors) | c = n/V, dilution and titration
Students drop cm3 straight into c = n/V, so every answer is ten or a thousand times out, and they average concentrations when diluting. This lesson drills two moves: convert volume first, then ask whether moles changed.
Objectives — Students can: (1) convert between mol, g and dm3 with c = n/V and convert cm3 to dm3 automatically; (2) use c1V1 = c2V2 for dilution and explain that adding water changes concentration but not moles; (3) run the three-step titration routine: n = cV, then the equation ratio, then the unknown concentration.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Mole and mass | molar mass of NaCl | 58.5 g/mol |
| Unit conversion | 250 cm3 in dm3 | 0.250 dm3 |
| Balancing equations | HCl + NaOH ratio | 1 : 1 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-10 min | Definition and units | c = n/V, V in dm3 | unit conversion drill |
| 10-24 min | Making a solution from mass | 5.85 g NaCl in 250 cm3 | 0.400 mol/dm3 |
| 24-34 min | Dilution keeps moles | 2.0 M, 25 cm3 to 250 cm3 | 0.200 M |
| 34-46 min | Three-step titration | 0.100 M HCl 20 cm3 vs 25 cm3 NaOH | 0.0800 M |
| 46-56 min | Preparing a target concentration | 0.2 M Na2CO3 in 500 cm3 | 10.6 g |
| 56-60 min | Exit check | three questions, marked at once | independent |
03 Teaching script (read it out)
Scene 1: One definition: moles per cubic decimetre
c = n/V with V in dm3, and 1 dm3 = 1000 cm3. 5.85 g NaCl gives n = 5.85 / 58.5 = 0.100 mol; dissolved to 250 cm3 = 0.250 dm3, so c = 0.400 mol/dm3.
Three details to say out loud: the volume is of the solution, not the solvent; the amount is in moles, not grams; mol/dm3 and M mean the same thing but exam answers should spell it out.
Working backwards: for 500 cm3 of 0.2 M Na2CO3, first n = 0.2 x 0.5 = 0.10 mol, then m = 0.10 x 106 = 10.6 g. The order is always moles first, grams second.
Three details to say out loud: the volume is of the solution, not the solvent; the amount is in moles, not grams; mol/dm3 and M mean the same thing but exam answers should spell it out.
Working backwards: for 500 cm3 of 0.2 M Na2CO3, first n = 0.2 x 0.5 = 0.10 mol, then m = 0.10 x 106 = 10.6 g. The order is always moles first, grams second.
Task: dissolve 4.0 g of NaOH and make up to 500 cm3. Find the concentration (0.20 mol/dm3) and write both steps.
Scene 2: Dilution: water changes concentration, not moles
Take 25 cm3 of 2.0 M solution: n = 2.0 x 0.025 = 0.050 mol. Make up to 250 cm3 and the moles stay at 0.050, so c = 0.050 / 0.25 = 0.200 mol/dm3.
So c1V1 = c2V2 is just conservation of moles, not a new formula. Volume up ten times, concentration down ten times; students who say it with their own numbers remember it.
Misconceptions to kill: averaging 2.0 M and water to get 1.0 M (only true for equal volumes by luck); and thinking dilution destroys solute. The squash-and-water image fixes this in ten seconds.
So c1V1 = c2V2 is just conservation of moles, not a new formula. Volume up ten times, concentration down ten times; students who say it with their own numbers remember it.
Misconceptions to kill: averaging 2.0 M and water to get 1.0 M (only true for equal volumes by luck); and thinking dilution destroys solute. The squash-and-water image fixes this in ten seconds.
Task: dilute 25 cm3 of 2.0 M to 250 cm3 and state whether the moles changed (0.200 M, unchanged).
Scene 3: Titration in three fixed steps
0.100 M HCl used at 20 cm3 gives n(HCl) = 0.100 x 0.020 = 0.0020 mol. HCl + NaOH gives NaCl + H2O, ratio 1 : 1, so n(NaOH) = 0.0020 mol. At 25 cm3 = 0.025 dm3, c(NaOH) = 0.0020 / 0.025 = 0.0800 mol/dm3.
Write the three steps in the same order every time, because examiners mark that structure. The ratio step costs the most marks: with H2SO4 + 2NaOH, n(NaOH) is twice n(H2SO4).
Sanity check: lab concentrations sit between 0.05 and 1 M. An answer of 8 M or 0.0008 M usually means cm3 were never converted.
Write the three steps in the same order every time, because examiners mark that structure. The ratio step costs the most marks: with H2SO4 + 2NaOH, n(NaOH) is twice n(H2SO4).
Sanity check: lab concentrations sit between 0.05 and 1 M. An answer of 8 M or 0.0008 M usually means cm3 were never converted.
Task: 25.0 cm3 of H2SO4 needs 20.0 cm3 of 0.100 M NaOH. Find the acid concentration (0.0400 M) and state the ratio used.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | Concentration | n = m/M, c = n/V | 0.400 M |
| Calculation | Dilution | c1V1 = c2V2 | 0.200 M |
| Calculation | Titration | n, ratio, then c | 0.0800 M |
| Calculation | Preparing a solution | n = cV, m = nM | 10.6 g |
Sanity check (teach this): Convert every volume to dm3 first: 1 cm3 = 0.001 dm3. An answer a thousand times too large almost always means this step was skipped. Expect lab concentrations of 0.1 to 1 M and titres of 20 to 30 cm3, so moles land near 10^-3 mol; an answer of 2 mol means the volume is still in cm3. Dilution must lower the concentration and keep the moles, so a rise in concentration means the formula is inverted.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| 5.85 g NaCl made up to 250 cm3: concentration? | 0.400 mol/dm3 | was cm3 converted |
| 25 cm3 of 2.0 M diluted to 250 cm3: new concentration? | 0.200 mol/dm3 | moles conserved |
| 25 cm3 NaOH needs 20 cm3 of 0.100 M HCl: find c(NaOH) | 0.0800 mol/dm3 | all three steps present |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Substituting cm3 as dm3 | 5.85 / 58.5 / 250 | 250 cm3 is 0.25 dm3; your units are three decimal places out. |
| Averaging concentrations on dilution | (2.0 + 0) / 2 = 1.0 M | Adding water cannot add solute, so use c1V1 = c2V2. |
| Skipping the equation ratio | treating H2SO4 as 1 : 1 | Write the equation first; a 2 in front must be multiplied, not guessed. |
| Using solvent volume as solution volume | dissolving in 250 cm3 of water and using 250 | Concentration refers to the final made-up volume. |
07 Homework (paper, section and question numbers)
- Secondary Chemistry chapter index — all chapters
- DSE Chemistry practice paper — solution and titration questions 4-9
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| What stays constant on dilution | moles of solute | correct: move to acids and bases |
| 250 cm3 in dm3 | 0.250 dm3 | wrong: reteach units |
| Titration step order | n, ratio, c | two or more wrong: back to act one |
09 Tutor FAQ
Why must volume be in dm3?
Because concentration is defined per dm3; mismatched units shift the answer by 1000 times, so converting first is safer than memorising formulas.
What happens to pH on dilution?
Diluting an acid tenfold raises pH by about 1, and diluting an alkali tenfold lowers it by about 1, but neither crosses 7.
Why repeat a titration three times?
Burette readings carry about 0.05 cm3 of error, so averaging reduces random error; results more than 0.20 cm3 apart point to a technique problem.
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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.
