Question

practice: density calculations
answer the following questions. make sure to show all work to receive credit. you may need a separate sheet of paper.

1. find the volume of a box measuring 2 cm by 7 cm by 3 cm.
2. an object placed in a graduated cylinder raises the volume from 12.2 ml to 14.5 ml. find the volume of the object.
3. find the volume of a cube measuring 5 cm on each side.
4. find the density of the box in #1 if it has a mass of 20 g.
5. find the density of the object in #2 if it has a mass of 4 g.
6. find the density of the cube in #3 if it has a mass of 100 g.
7. find the mass of an object that has a density of 1.5 g/cm³ and has a volume of 8 cm³
8. find the volume of an object with a density of 3.1 g/ml and a mass of 12 g.
9. find the mass of a cube that has a density of 2.7 g/ml and measures 3 cm on each side.
10. challenge: find the density of a soda can with a radius of 3.25 cm, a height of 12.2 cm, and a mass of 40 g.

Explanation:

Step1: Volume of a rectangular box

The formula for the volume \(V\) of a rectangular box is \(V = l\times w\times h\). Given \(l = 2\space cm\), \(w=7\space cm\), \(h = 3\space cm\), then \(V=2\times7\times3\).

$$V=2\times7\times3 = 42\space cm^{3}$$

Step2: Volume of an object using water - displacement

The volume of the object is equal to the change in volume of water in the graduated cylinder. Using the formula \(V=V_{final}-V_{initial}\). Given \(V_{final}=14.5\space mL\) and \(V_{initial} = 12.2\space mL\), then \(V=14.5 - 12.2\).

$$V=14.5- 12.2=2.3\space mL$$

Step3: Volume of a cube

The formula for the volume \(V\) of a cube is \(V = s^{3}\) (where \(s\) is the side - length). Given \(s = 5\space cm\), then \(V=5\times5\times5\).

$$V=5\times5\times5=125\space cm^{3}$$

Step4: Density of the box in #1

The formula for density \(
ho=\frac{m}{V}\). Given \(m = 20\space g\) and \(V = 42\space cm^{3}\) (from step1), then \(
ho=\frac{20}{42}\).

$$ ho=\frac{20}{42}\approx0.476\space g/cm^{3}$$

Step5: Density of the object in #2

Using the formula \(
ho=\frac{m}{V}\). Given \(m = 4\space g\) and \(V = 2.3\space mL\) (from step2), then \(
ho=\frac{4}{2.3}\).

$$ ho=\frac{4}{2.3}\approx1.74\space g/mL$$

Step6: Density of the cube in #3

Using the formula \(
ho=\frac{m}{V}\). Given \(m = 100\space g\) and \(V = 125\space cm^{3}\) (from step3), then \(
ho=\frac{100}{125}\).

$$ ho=\frac{100}{125}=0.8\space g/cm^{3}$$

Step7: Mass of an object

Using the formula \(m=
ho\times V\). Given \(
ho = 1.5\space g/cm^{3}\) and \(V = 8\space cm^{3}\), then \(m=1.5\times8\).

$$m=1.5\times8 = 12\space g$$

Step8: Volume of an object

Using the formula \(V=\frac{m}{
ho}\). Given \(m = 12\space g\) and \(
ho=3.1\space g/mL\), then \(V=\frac{12}{3.1}\).

$$V=\frac{12}{3.1}\approx3.87\space mL$$

Step9: Mass of a cube

First, find the volume of the cube. For a cube with \(s = 3\space cm\), \(V=s^{3}=3\times3\times3 = 27\space cm^{3}\). Then, using the formula \(m=
ho\times V\). Given \(
ho = 2.7\space g/mL=2.7\space g/cm^{3}\) (since \(1\space mL = 1\space cm^{3}\)), then \(m=2.7\times27\).

$$m=2.7\times27 = 72.9\space g$$

Step10: Density of a soda can

First, find the volume of the soda can (cylinder). The formula for the volume of a cylinder is \(V=\pi r^{2}h\). Given \(r = 3.25\space cm\) and \(h=12.2\space cm\), then \(V=\pi\times(3.25)^{2}\times12.2\).

$$V=\pi\times10.5625\times12.2\approx399.7\space cm^{3}$$

Then, using the formula \(
ho=\frac{m}{V}\). Given \(m = 40\space g\) and \(V\approx399.7\space cm^{3}\), then \(
ho=\frac{40}{399.7}\).

$$ ho=\frac{40}{399.7}\approx0.10\space g/cm^{3}$$

Answer:

1. \(42\space cm^{3}\)
2. \(2.3\space mL\)
3. \(125\space cm^{3}\)
4. \(0.476\space g/cm^{3}\)
5. \(1.74\space g/mL\)
6. \(0.8\space g/cm^{3}\)
7. \(12\space g\)
8. \(3.87\space mL\)
9. \(72.9\space g\)
10. \(0.10\space g/cm^{3}\)