Question

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 108cm and density is 10.5 g cm3, calculate the atomic mass of silver

Answer 1

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}

Plug the value we get

M = \frac{{10.5{{(4.07)}^3}6.022x{{10}^{23}}}}{4}

M = \frac{{10.5x67.41x6.022}}{4}

M = 107.09g{\rm{ }}mo{l^{ - 1}}