Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10^–8 cm and density is 10.5 g cm^–3, calculate the atomic mass of silver.

Given edge length (a) = 4.07 × 10 ^-8 cm

Density (d) = 10.5gcm^-1

For fcc number of atoms per unit cell (z) = 4

Avagadro number N_A = 6.022 × 10²³ _mol-1

We know that, Atomic mass (M)

(M)

Thus, M

⇒ M = 106.5 mol^-1