Density & Floating — Practice · Grade 8 Science

Chapter: Properties of Matter

What this practice covers

By the end of Density & Floating you should be able to:

11 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: density, displacement, average density, model, evidence, system boundary, assumption.

Problem 1

A block has mass 24 g and volume 8 cm³. Find its density.
Answer and explanation

3 g/cm³

24 ÷ 8 = 3.

Problem 2

Which measurement is needed with mass to calculate density?
  1. Time
  2. Volume
  3. Speed
  4. Force
Answer and explanation

Volume

Density = mass ÷ volume.

Problem 3

Why can a steel ship float?
Answer and explanation

Its hull encloses air, reducing its average density below that of water.

Problem 4

A solid has mass 54 g and raises water from 20 mL to 40 mL. Find density and predict its behavior in water of density 1 g/mL. Explain the reasoning in your own words.
Answer and explanation

2.7 g/mL; the solid sinks if fully wetted, without trapped air or other support.

Displaced volume is 40−20=20 mL.

Density is 54/20=2.7 g/mL, greater than water's density.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 5

An insoluble solid has mass 35 g and displaces 14 mL of water. Find its density and show the volume method. Show the reasoning, not just the conclusion.
Hints

Multiply density by volume: 3×8=24 and 3×24=72, recovering the masses.

Answer and explanation

The solid volume is 14 mL; density=35/14=2.5 g/mL.

Problem 6

A hollow object has total mass 90 g and external volume 120 mL. Predict floating in water at 1 g/mL using its average density.
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

90/120=0.75 g/mL, below water's density. It can float under the model, provided it retains its enclosed air and does not take in water.

Problem 7

Why does B not have twice A’s density?
Answer and explanation

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³.

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 8

Illustrative block A has mass 24 g and volume 8 cm³; B has mass 48 g and volume 16 cm³. A separate hollow object has mass 48 g and enclosed volume 80 cm³. Calculate the three average densities and use modeled water density 1 g/cm³ to predict floating. Explain why mass alone does not rank floating behavior.
Answer and explanation

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 9

An illustrative unknown has mass 54 g and displacement 20 cm³. References A/C are both 2.7 g/cm³ and B is 7.9. Argue which candidates remain and plan one additional property comparison.
Hints

Use the illustrated study titled Density as identification evidence. Keep the observations, model and conditions separate.

Answer and explanation

Density is 54/20=2.7 g/cm³. A and C remain consistent while B is inconsistent under matched conditions and adequate measurement precision. Density alone cannot distinguish A from C; choose a safe characteristic-property comparison with supplied reference data, such as solubility for approved samples. Name uncertainty and whether the object is homogeneous.

Density is 54/20=2.7 g/cm³. A and C remain consistent while B is inconsistent under matched conditions and adequate measurement precision. Density alone cannot distinguish A from C; choose a safe characteristic-property comparison with supplied reference data, such as solubility for approved samples. Name uncertainty and whether the object is homogeneous.

Response criteria

  • Calculate with correct units.
  • Distinguish support from unique identification.
  • Choose relevant further evidence and controls.

Problem 10

Measure an irregular object by displacementMeasure an irregular object by displacementBefore: 40 mLAfter: 58 mLObject volume = displaced liquid volume = 58 − 40 = 18 mL.1 mL = 1 cm³; read the scale and use complete submersion.
The two graduated-cylinder records show a rise from 40 to 58 mL when an object is fully submerged without trapped bubbles or spilled liquid. Its displaced volume is 18 mL, equivalent to 18 cm³. The cylinders are schematic: use the written readings rather than measuring the artwork, and do not confuse final total volume with object volume.
For this unit’s illustrated case, if the object has mass 54 g, what density follows from these readings?
Answer and explanation

Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.

First find the change in liquid reading; it is the volume of the submerged object. Then combine that volume with a separately measured mass. A water displacement of 18 mL does not mean the object itself has a mass of 18 g.

Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 11

A completely submerged object raises a cylinder reading from 35 to 47 mL and has mass 30 g. Calculate density, state the volume assumptions and distinguish this from claiming an exact substance identity.
Answer and explanation

Volume 12 cm³; density 30/12 = 2.5 g/cm³. Assume no trapped bubbles, spill or incomplete submersion. Density alone does not uniquely identify the substance.

Use the change 47−35, not the final reading.

Divide the independently measured mass by that volume.

Volume 12 cm³; density 30/12 = 2.5 g/cm³. Assume no trapped bubbles, spill or incomplete submersion. Density alone does not uniquely identify the substance.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

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