A metal forms fcc unit cell with unit edge length 400 pm . Calculate the number of atoms present in 0.6 g of metal if the density of metal is 12.0 g/cm ^3

Step-by-step solution

Edge length of the unit cell, a = 400 pm = 4 10⁻⁸ cm Volume of the unit cell, V = a^3 = (4 10⁻⁸ cm )^3 = 64 10⁻²⁴ cm ^3 Mass of one unit cell = Volume Density = 64 10⁻²⁴ cm ^3 12.0 g/cm ^3 = 768 10⁻²⁴ g Number of unit cells in 0.6 g of metal = 0.6 768 10⁻²⁴ = 600 768 10²¹ = 0.78125 10²¹ Since the metal forms an fcc unit cell, the number of atoms per unit cell is 4 . Total number of atoms = 4 0.78125 10²¹ = 3.125 10²¹ Answer: 3.125 10²¹