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

Since , the lattice is ccp, the number of silver atoms per unit cell `= Z = 4`
Atomic 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, `(rho) = (Z_(eff) xx Aw)/(a^(3) xx (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 g cm^(-3)`