Density

Density links mass and volume in one equation, and it explains why objects float or sink. Examiners like it because one question can test a calculation, a practical method and unit conversion: three skills, four marks, frequent casualties.

What is density and how do you calculate it?

Density is the mass per unit volume of a substance. In words: density equals mass divided by volume. In symbols: $\rho = \dfrac{m}{V}$.

Quantity	Symbol	Unit (SI)	Common alternative
Density	$\rho$	kg/m³	g/cm³
Mass	$m$	kg	g
Volume	$V$	m³	cm³

The two unit systems must never mix. Water has a density of 1000 kg/m³, which is the same as 1.0 g/cm³. The conversion: $1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$. An object floats in a liquid when its density is less than the liquid’s density, and sinks when it is greater. That single sentence answers most floating questions on Core papers.

How do you measure the density of an irregular object?

Use the displacement method. Measure the mass on a balance. Part-fill a measuring cylinder with water and record the volume. Lower the object in gently with thread and record the new volume. The volume of the object equals the rise in reading. Then apply $\rho = \dfrac{m}{V}$. For a regular block, measure the sides with a rule and use $V = \text{length} \times \text{width} \times \text{height}$ instead. For liquids, weigh an empty beaker, add a measured volume, reweigh, and divide the mass difference by the volume.

Worked Exam Question

A stone has a mass of 81 g. A measuring cylinder contains 50 cm³ of water. When the stone is fully submerged, the reading rises to 80 cm³. (a) Calculate the volume of the stone. [1] (b) Calculate the density of the stone in g/cm³. [2] (c) The density of water is 1.0 g/cm³. Explain why the stone sinks. [1]

Solution (a). $V = 80 - 50 = 30\ \text{cm}^3$.

Solution (b). Equation: $\rho = \dfrac{m}{V}$. Substitute: $\rho = 81 \div 30$. Answer: $\rho = 2.7\ \text{g/cm}^3$.

Solution (c). The stone’s density ($2.7\ \text{g/cm}^3$) is greater than the density of water, so it sinks.

Mark scheme

• B1 (a): 30 cm³.
• M1 (b): $\rho = \dfrac{m}{V}$ or $81/30$ seen. A1: $2.7\ \text{g/cm}^3$ with unit.
• B1 (c): density of stone greater than density of water (comparison required).

Common Mistakes

• Using the final cylinder reading as the volume. 80 cm³ gives $\rho = 1.0\ \text{g/cm}^3$, which is wrong. Fix: subtract the initial reading first.
• Mixing grams with m³. Fix: work entirely in g and cm³, or entirely in kg and m³.
• Inverting the equation. $V/m$ gives $0.37$, which should look suspicious for stone. Fix: density of solids is usually above 1 g/cm³, so sanity-check the size.
• Explaining sinking by weight alone. “It is heavy” scores zero. Fix: compare densities explicitly.
• Forgetting that $\rho$ is the Greek letter rho, not $p$. Fix: write it clearly so it cannot be confused with pressure.

Exam Technique Tip

In any density calculation, write the units beside every number as you substitute: $\rho = 81\ \text{g} \div 30\ \text{cm}^3$. If the units on the page do not match (g with m³, for example), convert before dividing. This habit catches the conversion errors that 0625 examiner reports list under density nearly every session.

How This Is Examined

Density is Core content and appears on every route. Papers 1 and 2 favour MCQs comparing densities or testing the floating rule. Papers 3 and 4 set 3-4 mark displacement calculations like the one above. Paper 6 is where density earns its keep: describing the displacement method, reading meniscus diagrams and suggesting accuracy improvements are regulars. Learn the method as numbered steps, because “describe how you would find the density of the stone” is worth 4-5 marks and each step scores separately.