Question

Ag forms $CCP$ lattice with edge length $408.6$pm calculate density of silver. $(Atomicmass=107.9\mu )$

Answer 1

$a=408.6,pm=408.6\times {10}^{-12}m=408.6\times {10}^{-10}cm$
$m=107.9$
$Z=4$ for CCP
mass of Ag=$\frac{107.9}{6.023\times {10}^{23}}=17.93\times {10}^{23}g$
$d=\frac{ZM}{a^{3}NA}$
$d=\frac{4\times 17.93\times {10}^{-23}}{(408.6\times {10}^{-10}{)}^{3}6.023\times {10}^{23}}$
$d=1.74\times {10}^{-23}$ $g/c{m}^3$