Calculate the number of atoms present in 1.58 g metal if it form bcc structure [ a^3=1.58 10⁻²² ~g ]

Step-by-step solution

For a body-centered cubic structure, the relationship between density ρ , lattice parameter a , and atomic mass M is given by ρ = 2 M N_A a^3 , where Z = 2 atoms per unit cell. Rearranging, ρ a^3 = 2M N_A , which represents the mass of two atoms. Given ρ a^3 = 1.58 10⁻²² g, this corresponds to the mass of 2 atoms. The mass per atom is 1.58 10⁻²² 2 = 0.79 10⁻²² g. To find the number of atoms in 1.58 g, divide the total mass by the mass per atom: 1.58 0.79 10⁻²² = 2 10²² atoms. Final answer: 2 10²² atoms, correspondi