A Compound AB has 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?

1. We have,  AB has rock salt structure (A : B :: 1 : 1).

2. That is,
F.C.C.structure (n = 4) and formula weight of AB is 6.023 y  g
closest distance A–B = y $\frac{1}{3}$ nm.

Therefore,          edge length of unit cell, a = 2(A+ + B–) = 2$\times \frac{y}{3}\times$10^​-9m

Therefore,          density of AB    =$$\begin{gathered} \frac{n\times mol.wt.}{N_A V} \\ =\frac{4\times 6.023\times y\times {10}^{-3}}{6.023\times {10}^{23}\times (2\times {\displaystyle \frac{y}{3}}{10}^{-9}{)}^3} \end{gathered}$$

                                                 =             5.0 kg m–3