Measure a density — answer key

1. DQ9 · Practice

Model answer. (a) V = 5.0 cm × 4.0 cm × 2.0 cm = 40 cm³. (b) ρ = m ÷ V = 108 g ÷ 40 cm³ = 2.7 g/cm³ (the density of aluminium).

Checked automatically

• (a) Volume of the block: 40 cm³ (within ±0.5 cm³)
  • Common wrong answer 11 : You added the sides. A volume multiplies them: 5.0 × 4.0 × 2.0.
• (b) Density of the metal: 2.7 g/cm³ (2 significant figures)
  • Common wrong answer 0.37 : You divided the volume by the mass. Density = mass ÷ volume: 108 ÷ 40.
  • Common wrong answer 9.8 : You divided by the sum of the sides. Use the volume, 40 cm³.

A full answer includes

• (a) I multiplied the three sides: 5.0 × 4.0 × 2.0 = 40 cm³.
• (b) I divided the mass by the volume: 108 g ÷ 40 cm³ = 2.7 g/cm³.

Also accept

• (a) 40 cm³; (b) 2.7 g/cm³.

Watch for

• “The volume is 5.0 + 4.0 + 2.0 = 11 cm³.” — Adding the sides gives a length, not a volume. Multiply them: 5.0 × 4.0 × 2.0.
• “Density = 40 ÷ 108 = 0.37 g/cm³.” — That is volume ÷ mass. Density is mass per unit volume: 108 ÷ 40.

2. DQ10 · Practice

Options as numbered on the pupil sheet: 1 — Read the new level; 2 — Weigh the dry stone on a balance; 3 — Divide the mass by the volume; 4 — Lower the stone on a thread until it is completely under water; 5 — Pour water into a measuring cylinder and read the level; 6 — Subtract the first reading from the second to find the volume.

Model answer. a weigh the dry stone; b pour water into a measuring cylinder and read the level (these two can swap); c lower the stone on a thread until it is completely under water; d read the new level; e subtract to find the volume; f divide the mass by the volume.

A full answer includes

• Weigh the stone while it is dry, before it goes into the water.
• Read the water level before and after the stone goes in.
• Find the volume as the rise in level, then divide the mass by the volume.

Also accept

• a, b, c, d, e, f (or b, a, c, d, e, f: the water can be read before the dry stone is weighed)

Watch for

• “Lower the stone in first, then weigh it.” — A stone taken out of water carries water with it, so it would weigh too much. Weigh it dry, before it goes in.
• “Read only the level after the stone is in.” — The level after includes the water that was there before. You need both readings to find the rise.

3. DQ11 · Practice

Model answer. (a) V = 47 cm³ − 35 cm³ = 12 cm³. (b) ρ = 30.0 g ÷ 12 cm³ = 2.5 g/cm³.

Checked automatically

• (a) Volume of the stone: 12 cm³ (within ±0.5 cm³)
  • Common wrong answer 47 : That is the final level. The stone’s volume is the rise: 47 − 35.
  • Common wrong answer 82 : You added the readings. Subtract them: 47 − 35.
• (b) Density of the stone: 2.5 g/cm³ (2 significant figures)
  • Common wrong answer 0.64 : You divided by the final level. Use the rise, 12 cm³.
  • Common wrong answer 0.40 : You divided the volume by the mass. Density = mass ÷ volume.

A full answer includes

• (a) I subtracted the levels: 47 − 35 = 12 cm³.
• (b) I divided 30.0 g by 12 cm³ to get 2.5 g/cm³.

Also accept

• (a) 12 cm³; (b) 2.5 g/cm³.

Watch for

• “The volume of the stone is 47 cm³.” — 47 cm³ is the water and the stone together. The stone is only the rise: 47 − 35.
• “Density = 30.0 ÷ 47 = 0.64 g/cm³.” — Use the stone’s own volume, the rise of 12 cm³, not the final reading.

4. DQ12 · Practice

Model answer. (a) m = 148.4 g − 92.0 g = 56.4 g. (b) ρ = 56.4 g ÷ 40 cm³ = 1.41 g/cm³, so 1.4 g/cm³. Pressing zero (tare) with the empty cylinder on the pan would have given 56.4 g directly.

Checked automatically

• (a) Mass of the syrup: 56.4 g (within ±0.05 g)
  • Common wrong answer 148.4 : That includes the cylinder. Subtract the empty reading, 92.0 g.
• (b) Density of the syrup: 1.4 g/cm³ (2 significant figures)
  • Common wrong answer 3.7 : You divided the full reading, cylinder included. Use 56.4 g.
  • Common wrong answer 0.71 : You divided the volume by the mass. Density = mass ÷ volume.

A full answer includes

• (a) I subtracted the empty reading: 148.4 − 92.0 = 56.4 g.
• (b) I divided 56.4 g by 40 cm³ to get 1.4 g/cm³.

Also accept

• (a) 56.4 g; (b) 1.4 g/cm³.

Watch for

• “The mass of the syrup is 148.4 g.” — 148.4 g includes the 92.0 g cylinder. Subtract it, or tare the balance before pouring.
• “Density = 148.4 ÷ 40 = 3.7 g/cm³.” — That counts the glass as syrup. Use the syrup alone: 56.4 g.

5. DQ13 · Practice

Model answer. m = ρ × V = 0.92 g/cm³ × 27.0 cm³ = 24.84 g, so 25 g to 2 significant figures.

Checked automatically

• Mass of the ice cube: 25 g (2 significant figures)
  • Common wrong answer 29 : You divided the volume by the density. m = ρ × V: 0.92 × 27.0.
  • Common wrong answer 0.034 : You divided the density by the volume. m = ρ × V.

A full answer includes

• I rearranged to m = ρ × V.
• I calculated 0.92 g/cm³ × 27.0 cm³ = 24.8 g, so 25 g.

Also accept

• 25 g (24.8 g).

Watch for

• “m = 27.0 ÷ 0.92 = 29 g.” — Dividing gives a mass bigger than the volume, as if ice were denser than water. Mass = density × volume.
• “m = 0.92 ÷ 27.0.” — That gives a tiny number with the wrong unit. Every cubic centimetre has 0.92 g, so 27.0 cm³ has 27.0 × 0.92 g.

6. DQ14 · Practice

Model answer. 920 kg/m³. A cubic metre is a million cubic centimetres, and 1000 g = 1 kg, so 1 g/cm³ = 1000 kg/m³: 0.92 × 1000 = 920 kg/m³.

Checked automatically

• Density of the oil: 920 kg/m³ (within ±1 kg/m³)
  • Common wrong answer 0.00092 : You divided by 1000. 1 g/cm³ = 1000 kg/m³, so multiply.
  • Common wrong answer 0.92 : The number changes with the unit: multiply by 1000.

A full answer includes

• I multiplied by 1000: 0.92 × 1000 = 920 kg/m³.

Also accept

• 920 kg/m³.

Watch for

• “0.00092 kg/m³.” — A cubic metre of oil is a big tank of it, so it has a big mass in kilograms. Multiply by 1000, don’t divide.
• “0.92 kg/m³: the number stays the same.” — Changing the units changes the number: 1 g/cm³ = 1000 kg/m³.

7. DQ15 · Practice

Model answer. a and b. An air bubble on the stone adds its own volume to the rise, and using the new level counts the water as well: both make the volume too big, so the density is too small. Weighing the stone wet makes the mass too big (density too big); splashing water out makes the rise too small (density too big); reading the bottom of the meniscus is the correct method.

A full answer includes

• I chose a and b.
• I explained that both make the measured volume too big, so mass ÷ volume comes out too small.

Also accept

• a, b.

Watch for

• “c, weighing it wet, makes the density too small.” — Water on the stone adds to the mass, so mass ÷ volume comes out too big, not too small.
• “e, splashing, makes the density too small.” — Splashed water is lost, so the level rises less and the volume looks too small: the density comes out too big.

8. DT3 · Apply

Model answer. (a) ρ = 230 g ÷ 85 cm³ = 2.7 g/cm³. (b) Under water the rock takes up the space that water was in. The can was already full to the spout, so all the water it displaces runs out: the volume collected equals the rock’s volume.

Checked automatically

• (a) Density of the rock: 2.7 g/cm³ (2 significant figures)
  • Common wrong answer 0.37 : You divided the volume by the mass. Density = mass ÷ volume.

A full answer includes

• (a) I divided 230 g by 85 cm³ to get 2.7 g/cm³.
• (b) I explained that the rock pushes aside its own volume of water, and because the can was full to the spout, exactly that volume runs out.

Also accept

• (a) 2.7 g/cm³; (b) the displaced water overflows from the full can.

Watch for

• “The water collected is the water left in the can.” — The water left in the can is not measured. The overflow is the water the rock pushed aside.
• “The can need not be full at the start.” — If it is not full, the rising water first fills the can up to the spout and less runs out, so the volume would be too small.

9. DT4 · Apply

Model answer. (a) V = 5.5 cm³ − 5.0 cm³ = 0.5 cm³; ρ = 9.7 g ÷ 0.5 cm³ = 19.4 g/cm³, so 19 g/cm³. (b) Gold: 19 is close to 19.3 and more than twice brass’s 8.5. (c) In a narrow cylinder a small volume makes a bigger change in height, so a 0.5 cm³ rise can be read; in a wide one it would barely show.

Checked automatically

• (a) Density of the ring: 19 g/cm³ (2 significant figures)
  • Common wrong answer 1.8 : You divided by the final level. The ring’s volume is the rise: 5.5 − 5.0 = 0.5 cm³.
  • Common wrong answer 0.052 : You divided the volume by the mass. Density = mass ÷ volume.

A full answer includes

• (a) V = 5.5 − 5.0 = 0.5 cm³, and 9.7 g ÷ 0.5 cm³ = 19 g/cm³.
• (b) I said gold, because 19 g/cm³ is close to 19.3 and far from 8.5.
• (c) I said a small volume gives only a tiny rise, which a narrow cylinder with fine marks can show.

Also accept

• (a) 19 g/cm³; (b) gold; (c) a small volume shows as a readable rise.

Watch for

• “Density = 9.7 ÷ 5.5 = 1.8 g/cm³.” — Use the ring’s own volume, the rise of 0.5 cm³, not the final level.
• “It is brass, because 19 is far from 8.5.” — Check again: 19 is close to gold’s 19.3 and far from brass’s 8.5.