Calculate the number of atoms present in 90 g of metal if it forms bcc structure. [ a^3 = 1.8 10⁻²² g ]

Step-by-step solution

Mass of one unit cell is given by a^3 = 1.8 10⁻²² g. For a body-centered cubic (bcc) structure, the number of atoms per unit cell is Z = 2 . Mass of 2 atoms = 1.8 10⁻²² g. Mass of 1 atom = 1.8 10⁻²² 2 = 0.9 10⁻²² g. Number of atoms in 90 g of the metal = 90 0.9 10⁻²² = 100 10²² = 1.0 10²⁴ . Answer: 1.0 10²⁴