Cell edge a = 5.0Å = 5.0 × 10 m = 5 × 10-8 cm
Density = 3.8 g/cm3
Molecular mass of FeO = 56 + 16 = 72 u
NA = 6.022 × 1023
d = \(\frac{ZM}{N_Aa^3}\)
Z = \(\frac{dN_Aa^3}{M}\)
= \(\frac{3.8\times6.022\times10^{23}}{72}\times\)(5 × 10-8)3
= 3.97 = 4
Each cell contain 4 FeO molecule every FeO molecule contain one Fe2+ and one O2- ion.
So no. of Fe2+ ion =4 no.of O2- ion = 4.