A metal having molar mass 96 g mol ⁻¹ crystallizes in a cubic lattice. If the edge length of the unit cell is 400 pm and the density of the metal is 5.0 g cm ⁻³ , the number of atoms present in one unit cell is ________. (Given: N_A = 6.0 10²³ mol ⁻¹ )

Step-by-step solution

Given: Molar mass, M = 96 g mol ⁻¹ Edge length, a = 400 pm = 4 10⁻⁸ cm Density, d = 5.0 g cm ⁻³ Avogadro's number, N_A = 6.0 10²³ mol ⁻¹ The density of a cubic unit cell is given by the formula: d = Z M N_A a^3 Rearranging for the number of atoms per unit cell ( Z ): Z = d N_A a^3 M Substituting the values: Z = 5.0 6.0 10²³ (4 10⁻⁸)^3 96 Z = 30 10²³ 64 10⁻²⁴ 96 Z = 30 6.4 96 = 192 96 = 2 The number of atoms present in one unit cell is 2 . Answer: 2