Iron oxide $FeO$ crystallises in a cubic lattice with a unit cell edge length of $5.0 \ \mathring{A}$. If the 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.
$N_{A} = 6.0 \times 10^{23} \ mol^{-1}$
Given: Molar mass of $Fe$ and $O$ is $56$ and $16 \ g \ mol^{-1}$ respectively.
$N_{A} = 6.0 \times 10^{23} \ mol^{-1}$
- A$2$
- B$6$
- C$8$
- D$4$
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