# The aluminum foil on a certain roll has a total area of 18.5 m^2 and a mass of 1275 g. Using a density of 2.7 g per cubic centimeter for aluminum, what is the thickness in millimeters of the aluminum foil?

Since aluminum has a density of $2.7 g / {\left(c m\right)}^{3}$, $1275 g$ of aluminum has a volume of $472.2 c {m}^{3}$. This is calculated by the formula linking density to mass and volume, $\mathrm{de} n s i t y = m a s s / v o l u m e$.
Given that $v o l u m e = a r e a \times \mathrm{de} p t h$, where volume of the aluminum foil is $472.2 c {m}^{3}$ and the area of the foil is ($18.5 \times {10}^{3}$)$c {m}^{2}$, it follows that the depth, or the thickness, of the foil should be equal to $v o l u m e / a r e a$.
The volume and the area should all be expressed in the same units, which is why I converted the ${m}^{2}$ to $c {m}^{2}$. As you require the thickness in $m m$, the volume and area should be converted to $m {m}^{3}$ and $m {m}^{2}$ respectively.
So you have an aluminum foil of volume ($472.2 \times {10}^{3}$)$m {m}^{3}$ and area ($18.5 \times {10}^{4}$)$m {m}^{2}$. You need the thickness of such a foil, so you substitute the values in the formula $\mathrm{de} p t h = v o l u m e / a r e a$, which follows as $\mathrm{de} p t h = \left(472.2 \times {10}^{3}\right) / \left(18.5 \times {10}^{4}\right)$ $\implies 2.55 m m$ $\approx 2.6 m m$