Question

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

Answer 1

The mass of aluminum with a volume of `6.3 \ cm^3` is `17 \ grams`.

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 = M / V`.

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

According to Elmhurst College, the density of aluminum is `(2.7 \ grams)/(mL)`.

In order to use this density with a volume in `cm^3`, we need to convert it into `(grams)/(cm^3)`. Fortunately, `cm^3` and `mL` are equal to each other, so the density of aluminum is `(2.7 \ grams)/(cm^3)`

The equation for density can be rearranged to solve for mass: `V * D = M`

To get the value of `M`, we can multiply `(2.7 \ grams)/(cm^3)` and `6.3 \ cm^3` together, and you are left with the mass of the aluminum.

`6.3 \ cm^3 * (2.7 \ grams)/(cm^3) = 17.01 \ grams`

The `cm^3` both cancel, and you are left with `grams`.

From there, we use significant figures to show the precise digits of our answer. Both `6.3 \ cm^3` and `(2.7 \ grams)/(cm^3)` have only 2 significant figures, so we round our answer to 2 significant figures as well.

`17 \ grams` of aluminum is the final answer.