Question

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)

Answer 1

`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)`