Question
Silver crystallises in an fcc lattice. The edge length of its unit cell is 4.077 x 10–8 cm and its density is 10.5 g cm-3. Calculate on this basis the atomic mass of silver. (NA = 6.02 x 1023 mol–1).
Solution
We have,
Edge of length of cell a = 4.07x10–8cm
Density p = 10.5 g /cm3
Number of atoms in unit cell of fcc lattice = 4
Avogadro number NA = 6.022x1023
By using formula,