An element of atomic mass 98.5 g/ mol occurs in fcc structure. If its unit cell edge length is 500 pm and its density is 5.22 g/ cm35.22 g/ cm3 . Calculate the value of Avogadro constant.
We are given the edge length as 500 pm. This can be written as: a = 500 × 10-8cma = 500 × 10-8cm Given the density is = 5.22 g/cm−3g/cm−3 Atomic mass given is 98.5 g/mol We are given that the unit cell is FCC, and for FCC the number of atoms is 4 (Z = 4). We know the formula of density is: d = Z × Ma3 × N0d = Z × Ma3 × N0 Putting the values in the formula, we get: 5.22 = 4 × 98.5/(500 × 10-8cm)3 × N05.22 = 4 × 98.5(500 × 10-8cm)3 × N0 N0 = 6.03 × 1023 mol-1
