Question

The density of $KBr$ is $11.9{\text{g cm}}^{-3}$. The edge length of unit cell is $400\text{pm}$. This shows that $KBr$ has:
(Given: Molecular mass of $KBr=119\text{u}$)

• A. Face centered cubic lattice
• B. Body centered cubic lattice
• C. Simple cubic lattice
• D. None of these

Answer 1

The correct option is A Face centered cubic lattice

Given,
$\text{Density,}\rho =11.9{\text{g cm}}^{-3}\text{Edge length (a)}=400\text{pm}=400\times {10}^{-10}\text{cm}$
By using the relation,
$\rho =\frac{z\times M}{N_A\times a^3}$
$M=\text{Molar mass of KBr}z=\text{number of atoms per unit cell}{N}_A=\text{Avogadro's number}$
Substituting the values, we get
$11.9=\frac{z\times 119}{6.022\times {10}^{23}\times (400\times {10}^{-10}{)}^3}$
$\Rightarrow z=3.84\approx 4$
We know that, $z=4$ is for Face centered cubic lattice