Ag crystallises in fcc lattice. What is the total number of tetrahedral voids present in $540 \mathrm{~g}$…
Ag crystallises in fcc lattice. What is the total number of tetrahedral voids present in $540 \mathrm{~g}$ of Ag metal? $\left(N_A=\right.$ Avagadro number; $\mathrm{Ag}$ atomic weight $=108 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$10 N_A$
$20 N_A$
$40 N_A$
$60 N_A$
Solution
Ag crystallises in fcc lattice. Each unit cell contains 4 atoms and 8 tetrahedral voids.
The mass of one $\mathrm{Ag}$-atom
$
=\frac{\text { Atomic mass }}{N_A}=\frac{108 \mathrm{~g} \mathrm{~mol}^{-1}}{N_A}
$
The mass of one unit cell will be $=4 \times \frac{108 \mathrm{~g} \mathrm{~mol}^{-1}}{N_A}$
Number of unit cells in $540 \mathrm{~g}$ of crystal
$
=\frac{540 \mathrm{~g} \times N_A}{4 \times 108}=1.25 N_A
$
Number of tetrahedral voids in 1.25 unit cell
$
=N_A \times 8 \times 1.25=10 N_A
$