Question

Silver crystalline in fcc lattice if edge length of the cell is 407pm. And density is 10.5 g cm-3 calculate the atomic mass of silver

Answer 1

Solution

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

Use formula

Density p = \frac{{ZM}}{{{a^3}{N_A}}}

Cross multiply we get

ZM = pa3NA

Divide by Z we get

M = \frac{{p{a^3}{N_A}}}{Z}