Density Word Problems PDF | PDF | Density | Volume - Free Printable

Let's solve each of these density word problems step by step using the formula:

$$
\text{Density} = \frac{\text{Mass}}{\text{Volume}} \quad \text{or} \quad d = \frac{m}{V}
$$

We’ll show all work and include units.

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1. What is the density of carbon dioxide gas if 0.196 g occupies a volume of 100 mL?

- Mass $ m = 0.196 $ g
- Volume $ V = 100 $ mL

$$
d = \frac{0.196\ \text{g}}{100\ \text{mL}} = 0.00196\ \text{g/mL}
$$

✔ Answer: $ \boxed{0.00196\ \text{g/mL}} $

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2. A block of wood 3.0 cm on each side and has a mass of 27 g. What is the density of this block?

- Side length = 3.0 cm → It’s a cube
- Volume $ V = \text{side}^3 = (3.0)^3 = 27\ \text{cm}^3 $
- Mass $ m = 27 $ g

$$
d = \frac{27\ \text{g}}{27\ \text{cm}^3} = 1.0\ \text{g/cm}^3
$$

✔ Answer: $ \boxed{1.0\ \text{g/cm}^3} $

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3. An irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 2.0 mL. The height of the water rose to 7.0 mL. If the mass of the stone was 25 g, what was its density?

- Initial water volume = 2.0 mL
- Final volume = 7.0 mL
- Volume of stone = $ 7.0 - 2.0 = 5.0 $ mL
- Mass = 25 g

$$
d = \frac{25\ \text{g}}{5.0\ \text{mL}} = 5.0\ \text{g/mL}
$$

✔ Answer: $ \boxed{5.0\ \text{g/mL}} $

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4. A 10.0 cm³ sample of copper has a mass of 89.6 g. What is the density of copper?

- $ m = 89.6 $ g
- $ V = 10.0\ \text{cm}^3 $

$$
d = \frac{89.6\ \text{g}}{10.0\ \text{cm}^3} = 8.96\ \text{g/cm}^3
$$

✔ Answer: $ \boxed{8.96\ \text{g/cm}^3} $

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5. Silver has a density of 10.5 g/cm³ and gold has a density of 19.3 g/cm³. Which would have a greater mass, 5 cm³ of silver or 5 cm³ of gold?

Use $ m = d \times V $

- Silver: $ m = 10.5\ \text{g/cm}^3 \times 5\ \text{cm}^3 = 52.5\ \text{g} $
- Gold: $ m = 19.3\ \text{g/cm}^3 \times 5\ \text{cm}^3 = 96.5\ \text{g} $

Gold has greater mass.

✔ Answer: $ \boxed{5\ \text{cm}^3\ \text{of gold}} $

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6. Five mL of ethanol has a mass of 3.9 g and 5.0 mL of benzene has a mass of 4.4 g. Which liquid is denser?

Calculate density for both:

- Ethanol: $ d = \frac{3.9\ \text{g}}{5.0\ \text{mL}} = 0.78\ \text{g/mL} $
- Benzene: $ d = \frac{4.4\ \text{g}}{5.0\ \text{mL}} = 0.88\ \text{g/mL} $

Benzene is denser.

✔ Answer: $ \boxed{\text{benzene}} $

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7. A sample of iron has the dimensions of 2 cm × 3 cm × 2 cm. If the mass of this rectangular-shaped object is 94 g, what is the density of iron?

- Volume $ V = 2 \times 3 \times 2 = 12\ \text{cm}^3 $
- Mass $ m = 94 $ g

$$
d = \frac{94\ \text{g}}{12\ \text{cm}^3} = 7.83\ \text{g/cm}^3
$$

✔ Answer: $ \boxed{7.83\ \text{g/cm}^3} $

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8. A rectangular solid of unknown density is 5 meters long, 2 meters high, and 4 meters wide. The mass of this solid is 300 grams. Given this information for this homogeneous material, calculate the density.

First, find volume in cubic meters, then convert to cm³ because mass is in grams.

- Volume $ V = 5 \times 2 \times 4 = 40\ \text{m}^3 $
- Convert to cm³: $ 1\ \text{m}^3 = 1,000,000\ \text{cm}^3 $
So $ 40\ \text{m}^3 = 40 \times 10^6 = 4.0 \times 10^7\ \text{cm}^3 $
- Mass = 300 g

$$
d = \frac{300\ \text{g}}{4.0 \times 10^7\ \text{cm}^3} = 7.5 \times 10^{-6}\ \text{g/cm}^3
$$

This is extremely low — likely a typo in units? But based on given info:

✔ Answer: $ \boxed{7.5 \times 10^{-6}\ \text{g/cm}^3} $

> ⚠️ Note: This is unrealistic for a solid (e.g., foam or aerogel), but mathematically correct.

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9. A rock occupies a volume of 20 cm³ and has a mass of 54 g. Find the density of this rock.

- $ m = 54 $ g
- $ V = 20\ \text{cm}^3 $

$$
d = \frac{54\ \text{g}}{20\ \text{cm}^3} = 2.7\ \text{g/cm}^3
$$

✔ Answer: $ \boxed{2.7\ \text{g/cm}^3} $

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10. A rock has a density of 4 g/mL and a mass of 16 g. Find the volume of the rock.

Use $ V = \frac{m}{d} $

- $ m = 16 $ g
- $ d = 4\ \text{g/mL} $

$$
V = \frac{16\ \text{g}}{4\ \text{g/mL}} = 4\ \text{mL}
$$

✔ Answer: $ \boxed{4\ \text{mL}} $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ 0.00196\ \text{g/mL} $ |
| 2 | $ 1.0\ \text{g/cm}^3 $ |
| 3 | $ 5.0\ \text{g/mL} $ |
| 4 | $ 8.96\ \text{g/cm}^3 $ |
| 5 | $ 5\ \text{cm}^3\ \text{of gold} $ |
| 6 | $ \text{benzene} $ |
| 7 | $ 7.83\ \text{g/cm}^3 $ |
| 8 | $ 7.5 \times 10^{-6}\ \text{g/cm}^3 $ |
| 9 | $ 2.7\ \text{g/cm}^3 $ |
| 10 | $ 4\ \text{mL} $ |

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