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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 $=108 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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The correct answer is:
$10 N_A$
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
$$
=N_A \times 8 \times 1.25=10 N_A
$$
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
$$
=N_A \times 8 \times 1.25=10 N_A
$$
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