Iron has a body centered cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g/ cm-37.874 g/ cm-3 . Use this information to calculate Avogadro’s number. (At. Mass of Fe = 55.845u).
We are given the edge length as 286.65 pm. This can be written as: a = 286.65 × 10-10cma = 286.65 × 10-10cm Given the density is = 7.874 g/ cm-37.874 g/ cm-3 Atomic mass of the iron is given as 55.845 u We are given that the unit cell is BCC, and for BCC the number of atoms is 2 (Z = 2). We know the formula of density is: d = Z × Ma3 × N0d = Z × Ma3 × N0 Putting the values in the formula, we get: 7.874 = 2 × 55.845(286.65 × 10-10cm)3 × N07.874 = 2 × 55.845/(286.65 × 10-10cm)3 × N0 N0 = 6.022 × 1023 mol-1
