Silver forms $ccp$ lattice and $X$-ray studies of its crystals show that the edge length of its unit cell is $408.6 \,pm$. Calculate the density of silver ( Atomic mass $= 107.9 \,u$ ).

Since the lattice is $ccp$, the number of silver atoms per unit cell is $z = 4$.
Molar mass of silver $(M)$ $= 107.9 \,g \,mol^{-1} = 107.9 \times 10^{-3} \,kg \,mol^{-1}$.
Edge length of unit cell $(a)$ $= 408.6 \,pm = 408.6 \times 10^{-12} \,m$.
Density $(d)$ is calculated using the formula: $d = \frac{z \cdot M}{a^3 \cdot N_A}$.
Substituting the values: $d = \frac{4 \times (107.9 \times 10^{-3} \,kg \,mol^{-1})}{(408.6 \times 10^{-12} \,m)^3 \times (6.022 \times 10^{23} \,mol^{-1})}$.
$d = \frac{431.6 \times 10^{-3}}{68.22 \times 10^{-30} \times 6.022 \times 10^{23}} \,kg \,m^{-3} = 10.5 \times 10^3 \,kg \,m^{-3} = 10.5 \,g \,cm^{-3}$.