Calculate the number of atoms present in 1 g of an element if it forms fcc unit cell structure. [ a ^3=6.8 10⁻²² ~g ]

Step-by-step solution

Given that for an fcc unit cell of an element, the product a^3 = 6.8 10⁻²² g. The density of a unit cell is expressed as = Z M N_A a^3 , where Z is the number of atoms per unit cell, M is the molar mass, N_A is Avogadro’s number, and a is the edge length. Substituting into the given product: ( Z M N_A a^3 ) a^3 = 6.8 10⁻²² which simplifies directly to Z M N_A = 6.8 10⁻²² . For an fcc structure, Z = 4 , so 4 M N_A = 6.8 10⁻²² . The number of atoms in 1 g of the element is given by 1 M N_A = 1 ( M N_A ) . From the eq