Silver crystallises in $fcc$ lattice. If edge length of the cell is $4.07 \times 10^{-8} \ cm$ and density is $10.5 \ g \ cm^{-3}$,calculate the atomic mass of silver.

(N/A) Given: Edge length,$a = 4.07 \times 10^{-8} \ cm$
Density,$d = 10.5 \ g \ cm^{-3}$
For $fcc$ lattice,the number of atoms per unit cell,$z = 4$
Avogadro's number,$N_A = 6.022 \times 10^{23} \ mol^{-1}$
Using the formula: $d = \frac{z M}{a^3 N_A}$
Rearranging for molar mass $M$: $M = \frac{d a^3 N_A}{z}$
$M = \frac{10.5 \ g \ cm^{-3} \times (4.07 \times 10^{-8} \ cm)^3 \times 6.022 \times 10^{23} \ mol^{-1}}{4}$
$M = \frac{10.5 \times 67.419 \times 10^{-24} \times 6.022 \times 10^{23}}{4}$
$M = \frac{426.37}{4} \approx 106.59 \ g \ mol^{-1}$
Thus,the atomic mass of silver is approximately $106.59 \ u$.