# What mass of aluminum has a volume of 6.3 cm^3?

Dec 13, 2015

The mass of aluminum with a volume of $6.3 \setminus c {m}^{3}$ is $17 \setminus g r a m s$.

#### Explanation:

We can figure out this out by using aluminum's density. Density is the ratio of the amount of mass that can be fit into a certain volume.

The equation for density is $D = \frac{M}{V}$.

$M$ stands for mass, $V$ stands for volume, and $D$ stands for density.

According to Elmhurst College, the density of aluminum is $\frac{2.7 \setminus g r a m s}{m L}$.

In order to use this density with a volume in $c {m}^{3}$, we need to convert it into $\frac{g r a m s}{c {m}^{3}}$. Fortunately, $c {m}^{3}$ and $m L$ are equal to each other, so the density of aluminum is $\frac{2.7 \setminus g r a m s}{c {m}^{3}}$

The equation for density can be rearranged to solve for mass: $V \cdot D = M$

To get the value of $M$, we can multiply $\frac{2.7 \setminus g r a m s}{c {m}^{3}}$ and $6.3 \setminus c {m}^{3}$ together, and you are left with the mass of the aluminum.

$6.3 \setminus c {m}^{3} \cdot \frac{2.7 \setminus g r a m s}{c {m}^{3}} = 17.01 \setminus g r a m s$

The $c {m}^{3}$ both cancel, and you are left with $g r a m s$.

From there, we use significant figures to show the precise digits of our answer. Both $6.3 \setminus c {m}^{3}$ and $\frac{2.7 \setminus g r a m s}{c {m}^{3}}$ have only 2 significant figures, so we round our answer to 2 significant figures as well.

$17 \setminus g r a m s$ of aluminum is the final answer.