Density

1 Plan and risk assessment

Density is the mass of each cubic centimetre of a material: density = mass ÷ volume. It depends on what the object is made of, not on how big it is, so it can tell a gold crown from a gold-plated fake.

ρ = m ÷ V

Prediction

2 Apparatus

• ✓Digital balance (resolution 0.1 g) Measures mass.
• ✓Metal block (cuboid) The regular object.
• ✓30 cm ruler (resolution 1 mm) Measures the block’s sides.
• ✓Vernier callipers (resolution 0.1 mm) Measures the block’s sides more precisely.
• ✓Model crown The irregular object: is it gold?
• ✓Thin cotton thread and thick string To lower the crown into the water.
• ✓Eureka (displacement) can on a brick Overflows by the crown’s volume.
• ✓Measuring cylinders: 25 cm³, 100 cm³ (×2) and 250 cm³ (resolution 0.5 cm³, 1 cm³, 2 cm³) Measure volumes of water and solution.
• ✓Sugar solution (about 150 cm³) The liquid.
• ✓Plastic tray and paper towels Catch spills.

3 Method: your choices

• Zero the balance: DoneA balance that does not read zero with nothing on it adds the same error to every mass (a zero error).
• Choose how to measure the block: Vernier callipers (0.1 mm)Resolution 0.1 mm, ten times finer than the ruler. One division is 0.2–0.5 % of a 2–5 cm side. Errors in all three dimensions contribute to the volume uncertainty.
• Choose how to find the crown’s volume: Eureka can, catching the overflowThe water that runs out of the spout is the crown’s volume, measured in whichever cylinder you choose. A little water stays in the spout each time.
• Choose a measuring cylinder for the overflow: 25 cm³ (0.5 cm³ divisions)Each reading to about ±0.25 cm³ (half a division): under 3 % of the crown’s volume.
• Tie on the crown: Thin cotton threadAdds only about 0.03 cm³ of its own volume: too small to notice.
• Put the water work in a traySafety: DoneSpilt water makes the floor slippery, and water must not get into the balance.
• Wait for the can to stop dripping: DoneDrips still coming out of a just-filled can are not displaced by the crown.
• Read the meniscus at eye level: DoneLooking down at the scale from above, the meniscus lines up with a higher mark (parallax).

Anomaly: No anomalous reading.

6 Analysis

1. Volume of the block

   volume = length × width × height

   Calculate the volume of the block.

   17.2 cm³ Right first time

2. Density of the block

   density = mass ÷ volume

   Calculate the density of the block, using your volume.

   7.83 g/cm³ Right first time

3. What is the block made of?

   Your block’s density is 7.83 g/cm³. Use the table to identify the metal.

   Iron Right first time

4. Volume of the crown

   mean = sum of the values ÷ number of values

   Calculate the mean volume of the crown.

   10.3 cm³ Right first time

5. Density of the crown

   density = mass ÷ volume

   Calculate the density of the crown.

   8.56 g/cm³ Right first time

6. Is the crown gold?

   Your crown’s density is 8.56 g/cm³. What is it most likely made of?

   Gold-plated brass (8.5 g/cm³) Right first time

7. Shape of the graph

   What does your graph show?

   Mass is directly proportional to volume: a straight line through the origin. Right first time

8. Gradient (the density of the solution)

   gradient = change in y ÷ change in x

   1.09 g/cm³ Right first time

9. In kg/m³

   1 g/cm³ = 1000 kg/m³

   Your solution’s density is 1.09 g/cm³. Give it in kg/m³.

   1090 kg/m³ Right first time

7 Conclusion and evaluation

Conclusion

Sources of error

Improvement

Repeatable or reproducible

9 Exam questions on your data

On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 30/30.

1. 01.1[1 mark]

   Write down the equation that links density (ρ), mass (m) and volume (V).

   Your answer

   density = mass divided by volume

   1/1 self-marked

   • density = mass ÷ volume, ρ = m ÷ V (or any correct rearrangement, in words or symbols)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: density = mass ÷ volume ρ = m ÷ V

2. 01.2[3 marks]

   Table 1 shows your measurements of the metal block.

   Table 1

   Mass in g	Length in cm	Width in cm	Height in cm
   135.0	4.00	2.80	1.54

   Calculate the density of the block.

   Your answer

   7.83

   Working: V = 4.00 × 2.80 × 1.54 = 17.248 cm³ ρ = 135.0 ÷ 17.248 ρ = 7.83 g/cm³

   3/3

   • Volume = 4.00 × 2.80 × 1.54 = 17.248 cm³
   • Substitution: ρ = 135.0 ÷ 17.248
   • ρ = 7.83 g/cm³

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: V = 4.00 × 2.80 × 1.54 = 17.248 cm³ ρ = 135.0 ÷ 17.248 ρ = 7.83 g/cm³

3. 01.3[2 marks]

   You measured the sides of the block with Vernier callipers (resolution 0.1 mm). Another student measured the same block with a 30 cm ruler (resolution 1 mm).

   Explain why the Vernier callipers give a more accurate value for the density of the block than the ruler.

   Your answer

   The ruler has a lower resolution (1 mm) than the Vernier callipers (0.1 mm), so the lengths measured with the ruler are less accurate and have a larger percentage uncertainty.

   2/2 self-marked

   • The ruler has a lower resolution (1 mm) than the callipers (0.1 mm)
   • So each side (and the volume) measured with the ruler is less accurate, with a larger percentage uncertainty

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The ruler has a lower resolution (1 mm) than the callipers (0.1 mm), so each side, and so the volume, is measured less accurately with a larger percentage uncertainty.

4. 01.4[2 marks]

   Another student forgot to zero the balance. With nothing on the pan it showed 0.3 g.

   Describe how the student should correct the masses they recorded.

   Your answer

   Record the reading with nothing on the balance, which is the zero error of 0.3 g, then subtract 0.3 g from each mass reading.

   2/2 self-marked

   • Record the zero error: the reading (0.3 g) with nothing on the balance
   • Subtract 0.3 g from each mass reading

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: Record the reading with nothing on the balance (the zero error, 0.3 g), then subtract 0.3 g from every mass they recorded.

5. 02.1[6 marks]

   A student wants to find the density of a small rock. The rock has an irregular shape.

   Describe a method the student could use to determine the density of the rock.

   Your answer

   1. Measure the mass of the rock on a balance. 2. Fill a displacement (eureka) can with water until it runs out of the spout, and wait until it stops dripping. 3. Put an empty measuring cylinder under the spout. 4. Tie the rock to thin thread and lower it carefully into the can until it is completely under water. 5. Collect all the water that comes out of the spout in the measuring cylinder. 6. Read the volume of water collected at eye level, at the bottom of the meniscus. This is the volume of the rock. 7. Repeat, refilling the can each time, and calculate a mean volume. 8. Calculate the density using density = mass ÷ volume.

   6/6 self-marked

   • Measure the mass using a balance (scales)
   • Fill a eureka (displacement) can until water just runs out of the spout
   • Or: put some water in a measuring cylinder and note its volume first
   • Place the rock in the water (lowered on thread) so it is fully submerged
   • Collect the displaced water in a measuring cylinder
   • Read the final volume; volume of rock = final − initial (or = volume of water displaced)
   • Density = mass ÷ volume
   • Read the meniscus at eye level; wait for dripping to stop; repeat and take a mean

   Self-marked: Level 3, 6 of 6 marks. You ticked 8 of 8 indicative points. This is your own judgement against the mark scheme.

   Model answer: 1. Measure the mass of the rock on a balance. 2. Fill a displacement (eureka) can with water until it runs out of the spout, and wait until it stops dripping. 3. Put an empty measuring cylinder under the spout. 4. Tie the rock to thin thread and lower it carefully into the can until it is completely under water. 5. Collect all the water that comes out of the spout in the measuring cylinder. 6. Read the volume of water collected at eye level, at the bottom of the meniscus. This is the volume of the rock. 7. Repeat, refilling the can each time, and calculate a mean volume. 8. Calculate the density using density = mass ÷ volume.

6. 02.2[1 mark]

   You lowered the crown into the water on thin cotton thread.

   Explain why thin thread is better than thick string for lowering the object into the water.

   Your answer

   Thin thread displaces less water so it would affect the volume measurement less than thick string.

   1/1 self-marked

   • Thin thread takes up less space under the water than thick string, so it adds less to the volume reading

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: Thin thread displaces less water, so it affects the volume measurement less than thick string.

7. 02.3[2 marks]

   You measured the volume of the crown three times.

   Give two reasons for repeating the measurement.

   Your answer

   So that a mean can be calculated, which reduces the effect of random errors. Anomalies can also be identified and removed.

   2/2 self-marked

   • A mean can be calculated
   • Which reduces the effect of random errors
   • Anomalies can be identified (and removed)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: So that a mean can be calculated, which reduces the effect of random errors, and so that anomalies can be identified.

8. 02.4[1 mark]

   A classmate measured the same crown with a eureka can and a 25 cm³ measuring cylinder. In one trial they forgot to refill the can to the spout.

   Table 2 shows the classmate’s results.

   Which trial is anomalous?

   Select one row.

   Table 2

   Trial	Volume of water collected in cm³
   1	10.5
   2	6.5
   3	10.5

   Your answer

   trial 2

   1/1

   • Trial 2: much smaller than the other two, which agree

   Full marks. Your answer meets all the points for this question. Trial 2 (6.5 cm³) is far from the other two, which agree with each other.

   Model answer: Trial 2. It is much less than the other two, which agree with each other.

9. 02.5[3 marks]

   Table 2 shows the classmate’s volumes for the same crown.

   The mass of the crown was 88.5 g.

   Table 2

   Trial	Volume of water collected in cm³
   1	10.5
   2	6.5
   3	10.5

   Use Table 2 to calculate the density of the crown.

   Do not use the anomalous result.

   Your answer

   8.43

   Working: Mean volume = (10.5 + 10.5) ÷ 2 = 10.5 cm³ ρ = 88.5 ÷ 10.5 ρ = 8.43 g/cm³

   3/3

   • Mean volume of the other two trials = 10.5 cm³
   • Substitution: ρ = 88.5 ÷ 10.5
   • ρ = 8.43 g/cm³

   Full marks. Your answer meets all the points for this question.

   Model answer: Mean volume = (10.5 + 10.5) ÷ 2 = 10.5 cm³ ρ = 88.5 ÷ 10.5 ρ = 8.43 g/cm³

10. 02.6[1 mark]

    The crown is sold as a "gold crown". The table gives the density of some materials it could be made from.

    Use your answer to the previous question. What is the crown most likely to be made of?

    Tick one box.

    • Pure gold (19.3 g/cm³)
    • 22-carat gold (17.7 g/cm³)
    • ✓ Brass with a thin gold plating (8.5 g/cm³) (your choice)
    • Aluminium with a thin gold plating (2.7 g/cm³)

    Your answer

    Brass with a thin gold plating (8.5 g/cm³)

    1/1

    • Brass with a thin gold plating (8.5 g/cm³)

    Full marks. Clear and precise.

    Model answer: Brass with a thin gold plating (8.5 g/cm³): its density is closest to 8.43 g/cm³.

11. 02.7[1 mark]

    You read the volumes on a 25 cm³ measuring cylinder, which has 0.5 cm³ divisions. Another student wrote one volume as 9.65 cm³.

    Explain why the volume should not be written to 2 decimal places.

    Your answer

    The resolution of the measuring cylinder is 0.5 cm³ so it cannot be used to measure to 2 decimal places.

    1/1 self-marked

    • The resolution of the measuring cylinder is 0.5 cm³, so it cannot measure to 2 decimal places

    Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

    Model answer: The resolution of the measuring cylinder is 0.5 cm³, so it cannot measure volumes to 2 decimal places.

12. 03.1[2 marks]

    You plotted the mass of the sugar solution (y-axis) against its volume (x-axis).

    Your line of best fit passes through the origin and the point where the volume is 80 cm³ and the mass is 86.9 g.

    Calculate the density of the sugar solution.

    Your answer

    1.09

    Working: ρ = gradient = 86.9 ÷ 80 ρ = 1.09 g/cm³

    2/2

    • ρ = 86.9 ÷ 80 (the gradient)
    • ρ = 1.09 g/cm³

    Full marks. Clear and precise.

    Model answer: ρ = gradient = 86.9 ÷ 80 ρ = 1.09 g/cm³

13. 03.2[1 mark]

    Give your answer to the previous question in kg/m³.

    1 g/cm³ = 1000 kg/m³

    Your answer

    1090

    Working: 1.09 g/cm³ × 1000 = 1090 kg/m³

    1/1

    • 1090 kg/m³ (allow their previous answer × 1000)

    Full marks. Your answer meets all the points for this question.

    Model answer: 1.09 g/cm³ × 1000 = 1090 kg/m³

14. 03.3[4 marks]

    A small cube of iron has sides of 24.70 mm.

    The density of iron is 7.9 × 10³ kg/m³.

    Calculate the mass of the cube in kilograms.

    Give your answer to 3 significant figures.

    Your answer

    0.119

    Working: side = 24.70 mm = 24.70 × 10^-3 m V = (24.70 × 10^-3)^3 = 15.1 × 10^-6 m³ m = ρ × V = 7.9 × 10^3 × 15.1 × 10^-6 m = 0.119 kg

    4/4

    • Side in metres: 24.70 × 10⁻³ m
    • V = (24.70 × 10⁻³)³ = 15.1 × 10⁻⁶ m³
    • m = 0.119 kg (credit a mass that follows correctly from their own V)
    • Answer given to 3 significant figures

    Full marks. Your answer meets all the points for this question.

    Model answer: side = 24.70 mm = 24.70 × 10⁻³ m V = (24.70 × 10⁻³)³ = 15.1 × 10⁻⁶ m³ m = ρ × V = 7.9 × 10³ × 15.1 × 10⁻⁶ m = 0.119 kg

Unseen data: the same skills, new apparatus

1. 01[2 marks]

   A student measures a different simulated metal block. Use density = mass ÷ volume and cuboid volume = length × width × height.

   Block measurements

   Mass in g	Length in cm	Width in cm	Height in cm
   112.0	3.07	2.41	1.72

   Calculate the density of the block.

   Your answer

   8.8

   Working: 3.07 × 2.41 × 1.72 = 12.7 112 / 12.7

   2/2

   • Volume = 12.7 cm³.
   • Density = 8.8 g/cm³.

   Full marks. Well done.

   Model answer: Volume = 3.07 × 2.41 × 1.72 = 12.7 cm³. Using the unrounded volume, density = 112 ÷ (3.07 × 2.41 × 1.72) = 8.8 g/cm³.

2. 02[2 marks]

   A different simulated irregular solid has mass 248.0 g. It is fully submerged in water without spilling any water. Use density = mass ÷ volume.

   Measuring-cylinder readings

   Before submerging the solid in cm³	With the solid submerged in cm³
   150	164

   Calculate the density of the irregular solid.

   Your answer

   17.7

   Working: 164 - 150 = 14 248 / 14

   2/2

   • Displaced volume = 14 cm³.
   • Density = 17.7 g/cm³.

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: The solid displaces 164 − 150 = 14 cm³ of water. Density = 248 ÷ 14 = 17.7 g/cm³.

3. 03[2 marks]

   In a new simulated investigation the same empty cylinder is used throughout. A student plots total mass of cylinder and liquid vertically against volume of liquid horizontally. A best-fit line is drawn without forcing it through the origin.

   Example measurements from the investigation

   Volume of liquid in cm³	Total mass in g
   0	101.0
   20	123.4
   63	171.7
   99	212.9

   Which TWO statements are correct? Tick (✓) two boxes.

   • ✓ The line’s gradient still estimates the liquid’s density: the fixed container mass cancels when taking a change in mass. (your choice)
   • ✓ Dividing one total mass reading by its liquid volume overestimates the density because that mass includes the cylinder. (your choice)
   • The line must pass through the origin because the same liquid is used.
   • The vertical intercept gives the liquid’s density in g/cm³.
   • Repeating the measurements and taking a mean removes the container mass from every total mass reading.

   Your answer

   The line’s gradient still estimates the liquid’s density: the fixed container mass cancels when taking a change in mass.; Dividing one total mass reading by its liquid volume overestimates the density because that mass includes the cylinder.

   2/2

   • The line’s gradient still estimates the liquid’s density: the fixed container mass cancels when taking a change in mass.
   • Dividing one total mass reading by its liquid volume overestimates the density because that mass includes the cylinder.

   Full marks. Clear and precise.

   Model answer: The fixed empty-cylinder mass is an offset. It cancels in a change in mass, so gradient = change in mass ÷ change in volume estimates density. A single total mass ÷ liquid volume still includes the container and is too large. The vertical intercept estimates the cylinder mass, in g.