Question

A compound AB has a rock salt type structure with $A:B=1:1$. The formula weight of $AB$ is $6.023y\text{amu}$ and the closest $A-B$ distance is $y1/3\text{nm}$. Calculate the density $in{\text{kg m}}^{-3}$ of lattice.

Answer 1

AB has rock salt $A:B::1:1$ structure, i.e., FCC structure (n=4) and formula weight of $BB$ is $6.023yamu$ having closest distance $A-B{y}^{1/3}nm$. Therefore edge length of unit cell $=2(A^{+}+B^{-})=2\times y^{1/3}\times {10}^{-9}m$
$\therefore \text{Density of AB}=\frac{n\times molwt.}{AV.no.\times V}=\frac{4\times 6.023\times {10}^{23}\times y\times {10}^{-3}}{6.023\times {10}^{23}\times (2y^{1/3}\times {10}^{-9}{)}^3}(mol.wt.inkg)=5.0{\text{kg m}}^{-3}$