A compound AB has a rock salt-type structure with A:B = 1:1. The formula weight of AB is 6.023 y amu and the closest A–B distance is y1/3 nm. Calculate the density of lattice (in Kg/m³).

AB has rock salt (A: B: 1:1) structure, i.e., fcc structure (n = 4) and formula weight of BB is 6.023 y amu having closest distance A–By^1/3nm. Therefore, edge length of unit cell
$\begin{array}{l}=2\left({\mathrm{A}}^{+}+{\mathrm{B}}^{-}\right)=2\times {\mathrm{y}}^{1/3}\times {10}^{-9}\mathrm{m}\\ \therefore \text{ Density of }\mathrm{AB}=\frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3}\\ =\frac{4\times 6.023\times \mathrm{y}\times {10}^{-3}}{6.023\times {10}^{23}\times {\left(2{\mathrm{y}}^{1/3}\times {10}^{-9}\right)}^3}\end{array}$
(Molecular weight in kg) 

= 5.0 kg m^–3