DSE Chemistry | Moles and Concentration Lesson-Prep Guide (for tutors) | three conversion routes, dilution, four common errors

2026-11-05

Tutor lesson-prep guides · Secondary Chemistry

Students divide whatever number they see by whatever number is left, so the mole never becomes a concept. This lesson builds one picture: the mole sits in the middle, and mass, particle count and gas volume all travel through it.

Objectives — Students can: (1) state that the mole is a counting unit linking mass, particle count and gas volume, and write the three routes m/M, N/N_A and V/24; (2) convert cm3 to dm3 before any concentration calculation and explain why the divisor is 1000; (3) use c1V1 = c2V2 for dilution and a 1:1 mole ratio for titration, then sanity-check that the answer is smaller than the stock solution.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Relative atomic mass	What is the relative molecular mass of CO2?	44 (12 + 16 x 2)
Unit conversion	How many dm3 is 250 cm3?	0.250 dm3, divide by 1000
Molar mass	What is the molar mass of NaCl?	58.5 g/mol (23 + 35.5)

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Prior-knowledge check	Hand out three quick items and check answers with units	Work alone, circle what they got wrong
8-20 min	Build the mole as a bridge	Draw three boxes: mass, particles, gas volume, with mol in the centre	Students fill in the three equations and units
20-32 min	Three conversion routes	Work 22 g of CO2 into moles, molecules and volume	Students repeat and check the order of magnitude
32-44 min	Concentration and the unit trap	Write 250 cm3 as 0.250 dm3, then get 0.4 mol/dm3	Students name the step where 400 comes from
44-56 min	Dilution and titration	Show c1V1 = c2V2 and a 1:1 titration, writing units throughout	Pairs swap and mark each other
56-60 min	Exit check	Collect three items, mark at once, tag the act each error came from	Work independently

03 Teaching script (read it out)

Scene 1: Act 1: the mole counts, it does not weigh

The deepest misconception is that the mole is a new way of writing grams. Start with a dozen eggs: a dozen is twelve, and it can be eggs or ping-pong balls because the point is the count. The mole works the same way: 1 mol is 6.02 x 10^23 particles, and the particles can be atoms, molecules or ions.
Draw the three boxes on the board: mass in grams on the left, particle count on the right, gas volume in dm3 below, with a circle in the middle holding mol. Each route has exactly one divisor: mass by molar mass, particle count by Avogadro, gas volume by 24. Then ask: can mass go straight to particle count? The answer is no, everything passes through the middle circle. Work 22 g of CO2: 22 / 44 = 0.5 mol, then 0.5 x 6.02 x 10^23 = 3.01 x 10^23 molecules, and 0.5 x 24 = 12 dm3 at room conditions. All three answers come from the same 0.5 mol.

Students repeat: the mole is a count, not a weight, and every route passes through the middle circle.

Scene 2: Act 2: convert cm3 to dm3 before you divide

Act two is purely about units. Students write c = n/V and drop 250 straight in, producing 0.0004 or 400. The fix is a written line first: 250 cm3 = 250 / 1000 = 0.250 dm3, with the definition stated out loud, one dm3 is one thousand cm3, rather than memorised.
Model 5.85 g of NaCl: molar mass 58.5, so 0.1 mol, and 0.1 / 0.250 = 0.4 mol/dm3. Then run the contrast: forgetting the thousand gives 400 mol/dm3, which would mean 23 kg of salt in one litre of water. Students must voice that impossibility themselves. Put the second misconception on the board: high concentration does not mean more moles, since 0.1 dm3 of 2 mol/dm3 holds only 0.2 mol, less than 1 dm3 of 0.4 mol/dm3.

Students repeat: volume goes in dm3, so divide by a thousand the moment you see cm3.

Scene 3: Act 3: dilution and titration are the same idea

Dilution adds water only, so the moles of solute stay fixed and only the volume changes: c1V1 = c2V2. Once students can say no solute was added, the equation needs no memorising.
Model it: 25.0 cm3 of 0.40 mol/dm3 diluted to 250 cm3 gives 0.40 x 25 / 250 = 0.040 mol/dm3. The concentration falls tenfold while the volume rises tenfold, so the product is unchanged. Titration uses the mole ratio: 25.0 cm3 of NaOH needs 20.0 cm3 of 0.10 mol/dm3 HCl. Moles of HCl are 0.10 x 20 / 1000 = 0.002 mol; NaOH reacts 1:1, so it is also 0.002 mol; divide by 0.025 dm3 for 0.080 mol/dm3. One operation per line, with the unit written every time.

Students repeat: dilution adds no solute, titration follows the mole ratio.

04 Question types and worked steps

Type	Example	Steps	Answer
Mass to moles	How many moles in 22 g of CO2?	1 write the formula 2 molar mass 44 3 mass divided by molar mass	0.5 mol
Moles to particles	How many molecules in 0.5 mol of CO2?	1 multiply by 6.02 x 10^23 2 state the unit as molecules	3.01 x 10^23 molecules
Moles to gas volume	What volume does 0.5 mol of gas occupy at room conditions?	1 multiply by 24 2 unit is dm3	12 dm3
Concentration, dilution, titration	25.0 cm3 of 0.40 mol/dm3 diluted to 250 cm3	1 convert to dm3 2 c1V1 = c2V2 3 check the answer is smaller	0.040 mol/dm3

Sanity check (teach this): After every concentration, ask whether the number is possible. 5.85 g of NaCl in 250 cm3 gives 0.4 mol/dm3. An answer of 400 means cm3 were never converted; 0.0004 means cm3 were treated as dm3 and divided twice. A diluted solution must be weaker than the stock, and a titration answer should sit in the same order of magnitude as the standard solution.

05 Independent practice (3 questions)

Question	Answer	What to watch
How many moles and how many molecules are in 22 g of CO2?	0.5 mol; 3.01 x 10^23 molecules	Did they compute 44 first, or divide 22 by something else
25.0 cm3 of 0.40 mol/dm3 NaCl is diluted to 250 cm3. Find the new concentration.	0.040 mol/dm3	Did they convert to dm3 and use c1V1 = c2V2
25.0 cm3 of NaOH needs 20.0 cm3 of 0.10 mol/dm3 HCl. Find the NaOH concentration.	0.080 mol/dm3	Is the 1:1 ratio written down, and is the unit mol/dm3

06 Common mistakes and what to say

Mistake	Student writes	Say this
Dropping cm3 in as dm3	c = 0.1 / 250 = 0.0004 mol/dm3	Volume in dm3 first: 250 divided by a thousand is 0.25.
Wrong molar mass	CO2 taken as 12 + 16 = 28	Oxygen is two atoms, so 16 is doubled, which gives 44.
Adding concentrations when diluting	New concentration comes out larger after adding water	Water only dilutes, the moles did not grow, so the concentration must fall.
Forgetting the thousand in titration	0.10 x 20 = 2 mol of HCl	20 cm3 is 0.020 dm3, not 20 dm3.

07 Homework (paper, section and question numbers)

Redo this chapter — Redo the three practice items and write the equation and unit beside each
Chemistry chapter index — Preview the acid-base titration chapter and note two unclear steps
Conversion map — Draw the three-box map once and paste it on the first page of the notes

08 Exit check (last 5 minutes)

Question	Answer	Decision
What kind of unit is the mole	A counting unit: 1 mol = 6.02 x 10^23 particles	Saying weight means replay Act 1
How many dm3 is 250 cm3	0.250 dm3	Answering 250 or 2.5 means replay Act 2
Does dilution raise or lower concentration	Lower, because the moles stay the same while volume grows	Saying higher means replay Act 3

09 Tutor FAQ

How many equations must be memorised?

Only the three conversion routes plus c1V1 = c2V2 and the mole ratio; everything else can be derived in the exam.

Is molar volume 24 or 22.4?

Use 24 dm3/mol at room conditions and 22.4 at standard conditions, and always state the conditions in the answer.

Can a calculator take 6.02 x 10^23?

Yes, enter it as 6.02E23 and give the final answer to three significant figures.

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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.