# How to calculate the average size of cubic crystals of KCl powder? [closed]

The total area of $$\pu{0.5 g}$$ sample of $$\ce{KCl}$$ powder is measured to be $$\pu{1620 cm2}$$. If the powder consists of tiny cubic crystals, what is the average size of these crystals?

I'm studying for a physical chemistry final and am not sure how to do this practice question.
I know that the density is $$\delta = \frac{m}{A},\label{one}\tag1$$ where $$m$$ is the mass and $$A$$ is the surface area, so do I just find $$A=\frac{m}{\delta}$$ to find the average "size"?
Or do I use \eqref{one} and then find out what the unit cell structure of $$\ce{KCl}$$ powder is and use that to estimate what the surface area of the crystal is?

For a cube, the surface area is given by the formula $$6a^2$$ where $$a$$ is the length of one side. The volume is given by $$a^3$$. Surface area per volume is $$\frac{6a^2}{a^3}= \frac{6}{a}.$$
We know the total surface area to be $$\pu{1620 cm^2}$$. We can calculate the total volume from the mass of the $$\ce{KCl}$$ and its density.
\begin{align} V &= \frac{m}{\rho}\\ \rho(\ce{KCl}) &= \pu{1.98 g//cm^3}\\ V &= \frac{\pu{0.5 g}}{\pu{1.95 g//cm^3}}\\ &= \pu{0.3 cm^3} \end{align}
\begin{align} \frac{6}{a} &= \frac{\pu{1650 cm^2}}{\pu{0.3 cm^3}}\\ a &= \frac{6 \times \pu{0.3 cm^3}}{\pu{1650 cm^2}}\\ a &= \pu{9E-4 cm} \end{align}