A certain metal with a molar mass of 48 g/mol crystallizes in a cubic lattice. The density of the metal is 20 g/cm ^3 and the edge length of its unit cell is 200 pm . The coordination number of the metal atom in this crystal lattice is ___ . (Given: Avogadro's number N_A = 6 10²³ mol ⁻¹ )

Step-by-step solution

The formula for the density of a cubic unit cell is: d = Z M N_A a^3 Given: d = 20 g/cm ^3 M = 48 g/mol N_A = 6 10²³ mol ⁻¹ a = 200 pm = 2 10⁻⁸ cm Volume of the unit cell, a^3 = (2 10⁻⁸)^3 = 8 10⁻²⁴ cm ^3 Rearranging the density formula to solve for the number of atoms per unit cell ( Z ): Z = d N_A a^3 M Z = 20 6 10²³ 8 10⁻²⁴ 48 Z = 20 48 10⁻¹ 48 = 2 Since Z = 2 , the metal crystallizes in a body-centered cubic (BCC) lattice. The coordination number of an atom in a BCC lattice is 8 . Answer: 8