An element with density 11.2 gm/cm³ forms a fcc lattice with edge length of 4 × 10⁻⁸ cm. Calculate the atomic mass of the element. (NA = 6.02 × 10²³ mol⁻¹)

Given that,
Density, d = 11.2 g/cm³
Edge length, a = 4 × 10⁻⁸ cm
Avogadro's number, NA = 6.02 × 10²³ /mol
Atomic mass, M = ??
For f.c.c lattice, number of atoms per unit cell, z = 4
Using the formula,
d = MzNAa³
M = dNAa³z
M = 11.2 × 6.023 × 10²³ × (4 × 10⁻⁸)³ / 4
M = 107.9 ≈ 108 g/mol
Hence, the atomic mass of the element is 108 g/mol. It is silver metal.