Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0 Ao. Assuming density of the oxide as 4.0 g cm-3, then the number of Fe2+ and O2- ions present in each unit cell will be

Let the units of ferrous oxide in a unit cell = n,molecular weight of ferrous oxide (FeO) = 56 + 16 = 72 g mol-1, Weight of n units =72×n6.023×1023Volume of one unit - (length of corner)3=(5Å)3=125×10−24cm3Density = wt.of cell  volume ,4.09=72×n6.023×1023×125×10−24n=3079.2×10−172=42.7×10−1=4.27≈4