Solid State Question 3
Question 3 - 30 January - Shift 2
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.
$N_A=6.0 \times 10^{23} mol^{-1}$
Show Answer
Answer: (4)
Solution:
$d=\frac{Z \times M}{N_0 \times a^{3}}$
$4=\frac{Z \times 72}{6 \times 10^{23} \times 125 \times 10^{-24}}$ $Z=4.166 \simeq 4$
