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IGCSE · Cambridge (CIE) · 0625

Rearrange density equations and convert cubic units

Find mass or volume and convert g/cm³ to kg/m³ using a cubic length conversion.

Density pathway · 1.4

Core and Extended; Physics 0625, 2026–2028, section 1.4. Density is mass per unit volume. Match the units before calculating. The syllabus includes measuring liquid, regular-solid and sinking irregular-solid densities, then using density data to predict floating. Paper exercises support practical reasoning but do not demonstrate supervised practical competence. Written practice is self-checked

Before you begin

  • Rearrange a division equation.
  • Cube the factor in a length-unit conversion.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

How many cm³ are in 1 m³?

What you will learn

  • Use m = ρV and V = m/ρ.
  • Convert density and volume units consistently.

Isolate the required quantity

From ρ = m/V, multiply by V to get m = ρV. A material of density 2.7 g/cm³ occupying 15 cm³ has mass 40.5 g. For mass 54 g at density 2.7 g/cm³, rearrange to V = m/ρ = 20 cm³.

Check units before substituting. Multiplying g/cm³ by cm³ leaves g. Dividing g by g/cm³ leaves cm³. If the volume is supplied in m³ while density is in g/cm³, convert one of them first rather than multiplying incompatible numerical values.

A cubic conversion uses the length factor three times

One metre is 100 cm, so one cubic metre contains 100³ = 1,000,000 cm³. Thus 1 cm³ = 10⁻⁶ m³. Also 1 g = 0.001 kg, so 1 g/cm³ = 0.001/10⁻⁶ = 1000 kg/m³. A density of 2.7 g/cm³ is 2700 kg/m³.

For a cross-check, 54 g becomes 0.054 kg and 20 cm³ becomes 0.000020 m³. Dividing gives 2700 kg/m³, the same density. Do not convert cm³ to m³ by dividing by only 100: that is a length conversion, not a volume conversion.

Pause and explain

What is 2.7 g/cm³ in kg/m³?

Put the idea to work

Worked example

Convert 2.7 g/cm³ to kg/m³ and find the volume of 54 g at that density.

Show the worked solution
  1. Multiply the density by 1000 to obtain 2700 kg/m³.
  2. Using the original matched units, V = 54/2.7.
  3. V = 20 cm³ = 0.000020 m³.

Answer 2700 kg/m³; 20 cm³ or 0.000020 m³.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Find the mass of 15 cm³ at density 2.7 g/cm³.

Give me a hint

Multiply density by volume with matching cubic-centimetre units.

Compare my reasoning
  1. Use m = ρV.
  2. 2.7 × 15 = 40.5.
  3. The remaining unit is g.

40.5 g.

Look for these in your work

  • Rearrange correctly.
  • Cancel volume units.
2 · Independent

Find the volume of 54 g at 2.7 g/cm³, then express both density and volume in SI units.

Give me a hint

Calculate in matched units, then apply the mass and cubic-volume factors.

Compare my reasoning
  1. V = 54/2.7 = 20 cm³.
  2. Density is 2700 kg/m³.
  3. Volume is 20 × 10⁻⁶ = 0.000020 m³.

20 cm³; 2700 kg/m³; 0.000020 m³.

Look for these in your work

  • Use m/ρ for volume.
  • Apply the cubic conversion.
3 · Transfer

A liquid density is 800 kg/m³ and its volume is 250 ml. Find mass in kg and g.

Give me a hint

One ml equals one cm³; convert the volume into m³.

Compare my reasoning
  1. 250 ml = 250 cm³ = 0.000250 m³.
  2. m = 800 × 0.000250 = 0.200 kg.
  3. Convert to 200 g.

0.200 kg; 200 g.

Look for these in your work

  • Convert the volume fully.
  • Retain consistent density units.

Common mistakes

  • Cube the length conversion for volume.
  • A density conversion combines mass and volume factors.
  • Use compatible units in m = ρV.
Recall without your notes

What can you explain now?

What mass occupies 15 cm³ at 2.7 g/cm³?

Compare with the explanation

40.5 g.

Multiply the density by the volume.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Derive the g/cm³ to kg/m³ factor from its two unit conversions without memorising it alone.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physics

Course scope and assessment

491 sequenced Physics lessons provide 1473 guided, independent and application problems. 58 of 58 syllabus sections have full linked teaching sequences; 0 sections remain pending. Three earlier starter overviews remain available but are not counted as full sequences. The map covers all six subject areas for 2026–2028 and keeps review and assessment limits visible. This is available content, not full programme completion, an exam grade or certified mastery. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Paper method exercises do not demonstrate practical competence or complete either practical-assessment route.

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