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 of Ag = 107.9 u)

Explanation

Since the lattice is ccp, the number of silver atoms per unit cell (Z) = 4
Molar mass of silver = 107.9 g `mol^(-1) = 107.9 xx 10^(-3) kg" "mol^(-1)`
Edge length of unit cell (a) = 408.6 pm = 408.6 `xx 10^(-12)` m
Density, d `= (Z.M)/(a^(3).N_(A))`
`= (4 xx (107.9 xx 10^(-3)kg" "mol^(-1)))/((408.6 xx 10^(-12) m)^(3) (6.022 xx 10^(23) mol^(-1)))`
`= 10.5 xx 10^(3) kg" "m^(-3)`
10.5 gm `cm^(-3)`