Iron (II) oxide has a cubic strcuture 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) = 6.02 xx 10^(23) "mol"^(-1)

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

1.	Iron (II) oxide has a cubic strcuture 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) = 6.02 xx 10^(23) "mol"^(-1)
Answer» <P> Solution :`Z= ( p xx a^(3) xx N_(0))/M ` `= ( ( 4 g cm^(-3))( 5 xx 10^(-8) cm)^(3)( 6.02xx 10^(23) "MOL"^(-1)))/( 72 "g mol" ^(-1)) = 4` Thus, there are 4 formula units (FEO)per unit cell. Hence, ` O^(2-) ` IONS = 4.