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Quick subject briefing
Before this lesson: Consolidation after pupils have met density and cuboid volume. They need a calculator. No measuring experiment is required today.
Density compares the mass in an equal volume. Dividing mass by volume gives the mass in one unit of volume. A smaller piece of the same uniform material has the same density: its mass and volume decrease together. Use matching mass and volume units before calculating.
Words and symbols you will need
- Density, ρ (rho)
- Mass per unit volume, expressed here in g/cm³ or kg/m³.
- Mass, m
- The amount of matter in the sample, measured here in g or kg.
- Volume, V
- The space occupied by the sample. A cuboid uses length × width × height.
- cm³ and m³
- Cubic centimetres and cubic metres: 1 cm³ = 0.000001 m³.
- Uniform material
- The material is the same throughout. Compare samples at the same temperature.
Help pupils choose a method
- Name the unknown: density = mass ÷ volume; mass = density × volume; volume = mass ÷ density.
- Match units: use g with cm³, or kg with m³. Divide cm³ by one million to obtain m³.
- Show the substitution and final unit. Compare material or liquid values in the same units.
Use the supplied model — pupil page 2
Pupil page 2 supplies the model. Multiply its three dimensions before dividing. Its 9 g in each cm³ becomes 9000 kg in each m³ because both the mass and volume units change.
A 54 g metal cuboid measures 3 cm × 2 cm × 1 cm. Calculate its density in kg/m³.
Worked answer: Volume = 3 × 2 × 1 = 6 cm³. Density = 54 ÷ 6 = 9 g/cm³. Convert: 9 × 1000 = 9000 kg/m³. Check using SI units: mass = 0.054 kg; volume = 0.000006 m³; density = 0.054 ÷ 0.000006 = 9000 kg/m³.
If pupils are still stuck
Read the relevant coaching aloud and use its named retry within the replacement time. If it remains insecure, help the pupil write the correct equation or first step, then record the exact difficulty and attempted working for the usual teacher.
KEEP THIS OPEN DURING THE LESSON
Timed teacher run sheet
- Download and print one complete seven-page pupil handout per pupil, in order. Double-sided printing is fine; staple each booklet.
- Read the subject briefing on teacher page 1. Use the timed run sheet on page 2 and the numbered answers. Each check has Say / Ask / Expect coaching and a fresh retry.
- Pupils need a pen, a different-coloured pen and a calculator. All reading, diagram and data are included; no pupil login, devices or practical apparatus are needed.
0–5 minPupil page / Q
1 / 1–3
Read aloud / check: “Start on page 1. Work from memory.” At 5 minutes check the three answers.
Next: If insecure, Try again A uses the first 2 reading minutes.
Answers: 1, 2, 3.
Open check 1 and its coaching →
5–15 minPupil page / Q
2 / reading
Read aloud / check: “Read the equations and conversions, then follow the worked model.”
Next: Check Try again A if used. The equations are supplied; focus on choosing and using them.
15–20 minPupil page / Q
3 / 4
Read aloud / check: “Choose a letter and show both conversions. Stop at the check.”
Next: Check B plus both conversions. If insecure, Try again B replaces 9–10.
Answers: 4.
Open check 2 and its coaching →
20–30 minPupil page / Q
3–4 / 5–10
Read aloud / check: “Work through the calculation questions. Show working and units.”
Next: If Try again B was used, leave 9–10 blank.
Answers: 5, 6, 7, 8, 9, 10.
30–33 minPupil page / Q
4 / 11
Read aloud / check: “Pause and answer question 11 independently.”
Next: Separate equation errors from cubic-unit errors. Try again C replaces 15–16.
Answers: 11.
Open check 3 and its coaching →
33–45 minPupil page / Q
5–6 / 12–21
Read aloud / check: “Use the cuboid and tables. Finish the comparisons.”
Next: Use the prepared data only. If finished early, strengthen the explanations in 14 and 16.
Answers: 12, 13, 14, 15, 16, 17, 18, 19, 20, 21.
45–55 minPupil page / Q
6 / 22
Read aloud / check: “Check your practice in a different colour. Explain one correction.”
Next: Read the numbered practice answers from this guide. Keep the exit answer hidden.
Answers: 22.
55–60 minPupil page / Q
7 / 23
Read aloud / check: “Cover the earlier pages. Complete the final question on your own.”
Next: Collect every booklet. Check the exit afterwards; D is next lesson only.
Answers: 23.
Open check 4 and its coaching →
Estimated route: 5 + 10 + 5 + 25 + 10 + 5 = 60 minutes. Practice is 10 minutes, a 3-minute check and 12 minutes. A uses reading time; B replaces 9–10; C replaces 15–16; D is next lesson only. Timings have not been classroom-trialled.
Check the entry, hinge and mid-practice responses at their marked pauses. At 45 minutes share only the practice answers. Keep the exit and next-lesson answers hidden until pupils submit.
Download teacher guide & answers ↓If several pupils are stuck
Use the coaching for the shared error, then the named independent retry within its replacement time. If the retry remains insecure, use the “If pupils are still stuck” guidance on teacher page 1 and record the specific difficulty for the usual teacher.
Estimated route: 5 + 10 + 5 + 25 + 10 + 5 = 60 minutes. Practice is 10 minutes, a 3-minute check and 12 minutes. A uses reading time; B replaces 9–10; C replaces 15–16; D is next lesson only. Timings have not been classroom-trialled.
ENTRY CHECK
Check 1: Question 1, Question 2, Question 3
Look for: Question 1: 180 g. Question 2: 24 cm³ from all three dimensions. Question 3: B, because equal volumes can be compared by mass.
If secure: Continue to the reading and model on pupil page 2.
If: Question 1 divides by 1000, or question 2 adds dimensions / uses only two.Open coaching
Likely issue: The pupil needs the direction of the mass conversion or the volume-of-a-cuboid rule.
- Say
- One kilogram is 1000 grams, so multiply kilograms by 1000. Cuboid volume multiplies three perpendicular lengths; adding lengths cannot give a volume.
- Ask
- How many grams are in 0.180 kg? What multiplication gives the volume of a 2 cm × 3 cm × 4 cm cuboid?
- Expect
- 180 g; 2 × 3 × 4 = 24 cm³.
Next: Use page 2: the mass-conversion row and the cuboid-volume sentence. Direct Try again A on page 1 within the first 2 reading minutes. Check all three new responses.
Pupil page 2: conversion table and Same material, different size; Try again A on page 1.
If: Question 3 gives A, or gives B without referring to the equal volumes.Open coaching
Likely issue: The mass-per-unit-volume comparison is insecure, or the correct choice may be a guess.
- Say
- Density compares mass in the same volume. When the volumes are equal, the sample with more mass has more mass per unit volume.
- Ask
- A and B each occupy 10 cm³. Which is denser: A at 20 g or B at 30 g? Why?
- Expect
- B: greater mass in the same volume. “B is heavier” alone needs the equal-volume reason.
Next: Read the opening definition on page 2. Ask what is held equal, then direct Try again A on page 1. Use reading time and check the independent comparison.
Pupil page 2, opening definition; Try again A on page 1.
HINGE CHECK
Check 2: Question 4
Look for: B, with 2.4 g = 0.0024 kg and 1 cm³ = 0.000001 m³, giving 2400 kg/m³. A correct letter alone does not demonstrate the method.
If secure: Start questions 5–10. If only the letter is correct, request both conversions before judging the method secure.
If: A leaves the value unchanged; C converts grams only.Open coaching
Likely issue: The density unit is a ratio: both numerator and denominator need conversion.
- Say
- Density is a mass divided by a volume. To change g/cm³ into kg/m³, change both units, then divide the converted mass by the converted volume.
- Ask
- For question 4, what do 2.4 g and 1 cm³ become in kg and m³?
- Expect
- 0.0024 kg and 0.000001 m³. Their ratio is 2400 kg/m³, option B.
Next: Use the g/cm³-to-kg/m³ row on page 2. Direct Try again B on page 3, replacing questions 9–10. Check both converted quantities and the final division.
Pupil page 2, final conversion row; Try again B on page 3.
If: D changes the volume unit but keeps grams as though they were kilograms, or uses 0.01 m³ for 1 cm³.Open coaching
Likely issue: One conversion has been omitted, or a length factor has been used for volume.
- Say
- A cubic centimetre has three lengths of 0.01 metres. Cube that length factor for volume. The mass must also change from grams to kilograms.
- Ask
- What is 0.01 × 0.01 × 0.01? What is 2.4 g in kilograms?
- Expect
- 0.000001 m³ for 1 cm³; 2.4 g = 0.0024 kg. Using 0.01 m³ converts only one length.
Next: Use “Convert volume as a cube” on page 2. Cube 0.01, then check the mass conversion. Direct Try again B instead of questions 9–10.
Pupil page 2, cubic-volume row and model; Try again B on page 3.
If: B with missing working or incompatible converted quantities.Open coaching
Likely issue: The choice may be guessed; the conversion method is not yet evidenced.
- Say
- A correct letter needs a checkable method. Write the mass in kilograms above the sample volume in cubic metres, then divide.
- Ask
- Talk through the two conversions and the division that give your answer to question 4.
- Expect
- 2.4 g = 0.0024 kg; 1 cm³ = 0.000001 m³; divide to obtain 2400 kg/m³.
Next: Ask the pupil to show the mass and cubic-volume conversions. If either is insecure, use the page 2 conversion rows and Try again B instead of questions 9–10.
Pupil pages 2–3, conversion rows and Try again B.
MID PRACTICE CHECK
Check 3: Question 11
Look for: 25 cm³ = 0.000025 m³; m = ρV = 1200 × 0.000025 = 0.030 kg. Check equation choice separately from conversion.
If secure: Continue with the cuboid and liquid applications, questions 12–21.
If: The volume conversion is correct, but divides density by volume or uses mass = volume/density.Open coaching
Likely issue: The requested quantity has not been connected to the correct rearrangement.
- Say
- Density tells us the mass in one unit of volume. To find the mass in several units of volume, multiply density by volume.
- Ask
- Which equation finds mass? Use it with the converted volume in question 11.
- Expect
- m = ρV; 1200 × 0.000025 = 0.030 kg. Dividing density by volume does not find mass.
Next: Use the “Mass, m” row on page 2. Direct Try again C on page 4 instead of questions 15–16. Require the equation before substitution.
Pupil page 2, mass equation; Try again C on page 4.
If: Uses m = ρV but substitutes 25, 0.25 or 0.025 as a volume in m³.Open coaching
Likely issue: The equation is appropriate; the cubic-unit conversion is the error.
- Say
- The equation is correct. The volume must be in cubic metres to match kg/m³. Divide a value in cm³ by one million.
- Ask
- Convert the 25 cm³ in question 11 to m³ before multiplying.
- Expect
- 25 ÷ 1 000 000 = 0.000025 m³. Then 1200 × 0.000025 = 0.030 kg.
Next: Use the cm³-to-m³ row on page 2. Ask the pupil to label both units, then complete Try again C instead of questions 15–16.
Pupil page 2, cubic-volume conversion; Try again C on page 4.
If: Equation and conversion are correct but the multiplication or final unit is wrong.Open coaching
Likely issue: This is an arithmetic/unit slip rather than a wrong density model.
- Say
- Keep the correct equation and volume conversion. Enter the whole decimal volume in the calculator. kg/m³ multiplied by m³ leaves kilograms.
- Ask
- What is 1200 × 0.000025, and which unit remains?
- Expect
- 0.030 kg. A correctly obtained 30 g must be converted to the requested kg.
Next: Keep the valid method. Check the calculator entry and that kg/m³ × m³ leaves kg. Complete Try again C independently instead of questions 15–16.
Pupil page 2, mass equation and units; Try again C on page 4.
EXIT CHECK
Check 4: Question 23
Look for: 8.0 g/cm³ = 8000 kg/m³. Each half: 48 g and 6 cm³, so density remains 8.0 g/cm³. Require the mass/volume ratio explanation.
If secure: Record the calculation, conversion and same-material comparison objectives as secure where the evidence supports them.
If: Gives 8 kg/m³, reverses the density equation or makes a cubic-unit error.Open coaching
Likely issue: Equation choice or conversion still needs an independent follow-up.
- Say
- Density is mass divided by volume. Grams divided by cm³ give g/cm³; multiply that density by 1000 to obtain kg/m³.
- Ask
- In question 23, what unit belongs to 96 ÷ 12? What is the same density in kg/m³?
- Expect
- 8.0 g/cm³, equivalent to 8000 kg/m³. The value 8 alone is not 8 kg/m³.
Next: Next lesson, revisit the equation/conversion tables and cuboid model on page 2. Then pupils cover the reading and complete Try again D on page 7. Compare with the key.
Pupil page 2, equations and conversions; Try again D on page 7, next lesson only.
If: Says each half is less dense because it has less mass, or gives unchanged density without a mass-and-volume reason.Open coaching
Likely issue: Sample size is being confused with the ratio defining density, or the reason is missing.
- Say
- Cutting the uniform sample halves both mass and volume. Density is their ratio, so it stays the same when both halve.
- Ask
- What are the mass and volume of each half in question 23? Divide them.
- Expect
- 48 g and 6 cm³; 48 ÷ 6 = 8.0 g/cm³, unchanged. Less mass alone does not establish less density.
Next: Next lesson, use “Same material, different size” on page 2. Halve mass and volume together, then direct Try again D independently.
Pupil page 2, Same material, different size; Try again D on page 7, next lesson only.
Answers keyed to the pupil handout
- Question 1
Pupil page 1 - 0.180 × 1000 = 180 g.
- Question 2
Pupil page 1 - V = 2 × 3 × 4 = 24 cm³. Multiply all three perpendicular dimensions.
- Question 3
Pupil page 1 - B. It has the greater mass in the same volume, so its mass per unit volume is greater.
- Question 4
Pupil page 3 - B. 2.4 g = 0.0024 kg; 1 cm³ = 0.000001 m³. Density = 0.0024 ÷ 0.000001 = 2400 kg/m³. Both mass and volume units change.
- Question 5
Pupil page 3 - Density: mass per unit volume, not total mass or weight.
- Question 6
Pupil page 3 - ρ = m/V = 54.0 ÷ 20.0 = 2.70 g/cm³.
- Question 7
Pupil page 3 - m = ρV = 2.70 × 35.0 = 94.5 g.
- Question 8
Pupil page 4 - V = m/ρ = 121.5 ÷ 2.70 = 45.0 cm³.
- Question 9
Pupil page 4 - 6.0 ÷ 1 000 000 = 0.0000060 m³, or 6.0 × 10⁻⁶ m³.
- Question 10
Pupil page 4 - 2.70 g = 0.00270 kg and 1 cm³ = 0.000001 m³. ρ = 0.00270 ÷ 0.000001 = 2700 kg/m³.
- Question 11
Pupil page 4 - 25 cm³ = 0.000025 m³. Use m = ρV: m = 1200 × 0.000025 = 0.030 kg. Accept 30 g only if correctly converted to the requested kg.
- Question 12
Pupil page 5 - V = 4.0 × 2.5 × 1.2 = 12 cm³.
- Question 13
Pupil page 5 - ρ = 106.8 ÷ 12 = 8.9 g/cm³. Use the pupil’s earlier valid volume for consistent follow-through.
- Question 14
Pupil page 5 - Copper: 8.9 g/cm³ matches the supplied value. Another material or alloy may have a similar density, so this is evidence rather than proof.
- Question 15
Pupil page 5 - ρ = 26.7 ÷ 3.0 = 8.9 g/cm³.
- Question 16
Pupil page 5 - Each has half the original mass and half the original volume. The ratio m/V is unchanged, so density is unchanged. Assume the same material and temperature.
- Question 17
Pupil page 6 - Liquid mass = 140.0 − 64.0 = 76.0 g.
- Question 18
Pupil page 6 - ρ = 76.0 ÷ 80.0 = 0.950 g/cm³.
- Question 19
Pupil page 6 - 0.950 × 1000 = 950 kg/m³.
- Question 20
Pupil page 6 - Liquid mass = 149.6 − 64.0 = 85.6 g. Density = 85.6 ÷ 80.0 = 1.07 g/cm³.
- Question 21
Pupil page 6 - B passes: 1.07 g/cm³ = 1070 kg/m³, within 1030–1090. A fails: 950 kg/m³ is below 1030. This is a density check, not proof of chemical suitability.
- Question 22
Pupil page 6 - Accept an accurate correction with a reason. For a unit error, explain that the centimetre-to-metre length factor is cubed for volume. For a rearrangement, use m = ρV or V = m/ρ as appropriate. For the all-correct option, mass and volume decrease in the same proportion, so m/V is unchanged.
- Question 23
Pupil page 7 - (a) ρ = 96 ÷ 12 = 8.0 g/cm³ = 8000 kg/m³. Alternatively 0.096 kg ÷ 0.000012 m³ = 8000 kg/m³. (b) Each half has mass 48 g and volume 6 cm³, so 48 ÷ 6 = 8.0 g/cm³: density is unchanged because mass and volume halve together.
- Try again A
- (a) 75 g. (b) 10 cm³. (c) The 25 g sample: more mass occupies the same volume.
- Try again B
- 1.6 g = 0.0016 kg; 1 cm³ = 0.000001 m³. Density = 0.0016 ÷ 0.000001 = 1600 kg/m³.
- Try again C
- V = 40 ÷ 1 000 000 = 0.000040 m³. m = ρV = 1500 × 0.000040 = 0.060 kg.
- Try again D
- 72 ÷ 30 = 2.4 g/cm³ = 2400 kg/m³. The quarter has mass 18 g and volume 7.5 cm³. Its density is still 2.4 g/cm³ because mass and volume both become one quarter.
Worksheet selection and marking
Accept equivalent working and correctly converted units. Award valid method/follow-through after arithmetic slips; do not confuse a missing unit with an inverted equation. Require a reason for comparisons. The 16 selected source parts are worth 21 practice marks before substitutions. Detailed density-measurement methods and practical completion are outside this lesson.
Selection: Source PDF pages 1–3 and 6–7: Q1(a–f), Q2(a–e), Q5(a–e). Sixteen parts, 21 practice marks before support substitutions.
Open the maintained worked solutions →Collect and follow up: Collect the complete named booklet. Use question 23 to record the objectives demonstrated and the precise next step: equation choice, cubic-unit conversion or unchanged density after cutting. Tell the usual teacher which pupils need Try again D.
