Question

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 $4.0gc{m}^{-3}$, then the number of $FeO$ units present per unit cell is ______ . (Nearest integer)
Given: Molar mass of $Fe$ and $O$ is $56$ and $16gmo{l}^{-1}$ respectively.
$N_A=6.0\times 10^{23}mo{l}^{-1}$

Answer 1

Answer: 4

$d=\frac{Z\times M}{N_0\times a^3}$
$4=\frac{Z\times 72}{6\times 10^{23}\times 125\times 10^{-24}}$
$Z=4.166\simeq 4$