# What is the density of a mineral with a mass of 412 g and a volume of 8.2 cm^3?

Jun 14, 2016

$50.2 \frac{g}{c {m}^{3}}$

#### Explanation:

In order to determine the density of the mineral, we have to use the following formula:

$D = \frac{M}{V}$

• D represents density and it will have units of $\frac{g}{m L}$ when talking about a liquid or units of $\frac{g}{c {m}^{3}}$ when talking about a solid.
• M represents mass and will have units of grams $g$.
• V represents volume and it will have units of $m L$ or $c {m}^{3}$

We know the mass and volume, both of which have the appropriate units so all we have to do is plug the numbers into the formula:

$D = \frac{412 g}{8.2 c {m}^{3}}$

$D = 50.2 \frac{g}{c {m}^{3}}$