Silver froms 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`).

Explanation

`rho=(ZxxM)/(a^3xxN_0)`
For ccp lattice (which is equivalent to fcc lattice ), Z=4 atoms/unit cell
Also we are given that `M=107.9 g mol^(-1)`
a=408.6 pm=`408.6xx10^(-12)` cm `therefore rho=(4 "atoms"xx107.9g mol^(-1)) /((408.6xx10^(-12)cm)^3 (6.022xx10^23 "atoms mol"^(-1))=10.5 g cm^(-3)`