Question

Question #543b6

Answer 1

The thing I want to point out is that density is defined as mass over unit of volume. Therefore, regardless of what actual unit you use to express it, you must always have a unit of mass divided by a unit of volume.

The density of aluminium should therefore be expressed in `"g/cm"^3`, since `"g/cm"` is actually a measure of mass per length, not per volume.

That being said, you go about solving this problem by using the formula for density

`rho = m/V => V = m/(rho)`

In this case,

`V_("cube") = m_("cube")/(rho) = "16.2 g"/(2.7 "g"/"cm"^3) = "6.0 cm"^3`

SInce you're dealing with a perfect cube, the volume can be expressed as

`V_("cube") = "side"^3`

Therefore,

`"side" = root(3)(6) = "1.82 cm"`