Question

Iron oxide $FeO$, crystallises in a cubic lattice with a unit cell edge length of $50 \mathring{A}$ If density of the $FeO$ in the crystal is $40\, g \,cm ^{-3}$, 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 $N _{ A }=60 \times 10^{23} mol ^{-1}$

Answer 1

The correct answer is 4.
d=N0​×a3Z×M​
4=6×1023×125×10−24Z×72​
Z=4.166≃4