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

`p= (Z xx M)/(a^(3) xxN_(0))`
For ccp lattic ( whichh 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 = ` 408.6 xx 10^(-12) cm therefore p = (4 "atoms" xx 107.9"g mol"^(-1))/((408.6 xx 10^(12)cm)^(3) (6.022 xx 10^(23) " atoms mol"^(-1))) = 10.5 " g cm" ^(-3)`