# Question 543b6

Feb 15, 2015

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 ${\text{g/cm}}^{3}$, since $\text{g/cm}$ 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

$\rho = \frac{m}{V} \implies V = \frac{m}{\rho}$

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,

$\text{side" = root(3)(6) = "1.82 cm}$