Calculate the number of atoms in 5.4 g metal forming fcc structure [ a^3=7 2 10⁻²² ~g ]

Step-by-step solution

Given a^3 = 7.2 10⁻²² g and FCC structure with Z = 4 atoms per unit cell, rearrange the density formula = ZM N_A a^3 to solve for molar mass: a^3 = ZM N_A . Substitute values using N_A = 6 10²³ mol ⁻¹ : 7.2 10⁻²² = 4M 6 10²³ . Solving yields M = 7.2 10⁻²² 6 10²³ 4 = 108 g/mol. The number of moles in 5.4 g is n = 5.4 108 = 0.05 mol. Thus the number of atoms is 0.05 6 10²³ = 3.0 10²² , corresponding to option A .