Iron (II) oxide has a cubic structure and each unit cell has side 5 Å. If the density of the oxide is 4 g cm^(-3),the number of oxide ions present in each unit cell is (Molar mass of FeO =72 "g mol"^(-1) , N_A=6.02xx10^23 mol^(-1))

Iron (II) oxide has a cubic structure and each unit cell has side 5 Å

1.	Iron (II) oxide has a cubic structure and each unit cell has side 5 Å. If the density of the oxide is 4 g cm^(-3),the number of oxide ions present in each unit cell is (Molar mass of FeO =72 "g mol"^(-1) , N_A=6.02xx10^23 mol^(-1))
Answer» Solution :`Z=(rhoxxa^3xxN_0)/M` `=((4 G CM^(-3))(5xx10^(-8)cm)^3(6.02xx10^23 MOL^(-1)))/(72 g mol^(-1))approx 4` Thus, there are 4 formula units (FeO) per unit cell.Hence, `O^(2-)` ions =4