Density. Problems worksheet - Free Printable

Let's solve the problems step by step using the formula for density:

Formula:

\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]

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Problem 1:

You have a mineral with a volume of \(15 \, \text{cm}^3\) and a mass of \(45 \, \text{g}\). What is its density?

#### Solution:
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{45 \, \text{g}}{15 \, \text{cm}^3} = 3 \, \text{g/cm}^3
\]

#### Answer:
\[
\boxed{3}
\]

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Problem 2:

You have a different mineral with a volume of \(15 \, \text{cm}^3\) and a mass of \(75 \, \text{g}\). What is its density?

#### Solution:
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{75 \, \text{g}}{15 \, \text{cm}^3} = 5 \, \text{g/cm}^3
\]

#### Answer:
\[
\boxed{5}
\]

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Problem 3:

In the above two examples, which mineral is more dense?

#### Solution:
From Problem 1, the density of the first mineral is \(3 \, \text{g/cm}^3\).
From Problem 2, the density of the second mineral is \(5 \, \text{g/cm}^3\).

Since \(5 \, \text{g/cm}^3 > 3 \, \text{g/cm}^3\), the second mineral is more dense.

#### Answer:
\[
\boxed{\text{second, first}}
\]

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Problem 4:

What is the density of a cube of sugar weighing \(12.8 \, \text{grams}\) measuring \(2 \, \text{cm}\) on a side?

#### Step 1: Calculate the volume of the cube.
The volume \(V\) of a cube is given by:
\[
V = \text{side}^3 = 2 \, \text{cm} \times 2 \, \text{cm} \times 2 \, \text{cm} = 8 \, \text{cm}^3
\]

#### Step 2: Use the density formula.
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{12.8 \, \text{g}}{8 \, \text{cm}^3} = 1.6 \, \text{g/cm}^3
\]

#### Answer:
\[
\boxed{1.6}
\]

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Final Answers:

1. \(\boxed{3}\)
2. \(\boxed{5}\)
3. \(\boxed{\text{second, first}}\)
4. \(\boxed{1.6}\)