Q.
Iron oxide $FeO$, crystallises in a cubic lattice with a unit cell edge length of $5.0 \mathring{A}$. If density of the $FeO$ in the crystal is $4.0\, g \,cm ^{-3}$, then the number of $FeO$ units present per unit cell is ______ . (Nearest integer)
Given: Molar mass of $Fe$ and $O$ is $56$ and $16 \, g \, mol ^{-1}$ respectively.
$N _{ A }=6.0 \times 10^{23} mol ^{-1}$
Solution: