Question

Silver forms $ccp$ lattice and $X$-ray studies of its crystals show that the edge length of its unit cell is $408.6pm$ The density of silver (Atomic mass $=107.9u)$ is

• A. $10.5gc{m}^{-3}$
• B. $12.5gc{m}^3$
• C. $5gc{m}^{-2}$
• D. $15.7gc{m}^{-3}$

Answer 1

Answer: (A)

Since the lattice is $ccp$ the number of silver atoms per unit cell $=Z=4$
Molar mass of silver $=107.9gmo{l}^{-1}=107.9\times 10^{-3}kgmo{l}^{-1}$
Edge length of unit cell $=a=408.6pm=408.6\times 10^{-12}m$
Density, $d=\frac{Z.M}{a^3\cdot N_A}$
$=\frac{4\times \left(107.9\times 10^{-3}kgmo{l}^{-1}\right)}{{\left(408.6\times 10^{-12}m\right)}^3\left(6.022\times 10^{23}mo{l}^{-1}\right)}$
$=10.5\times 10^{3}kg{m}^{-3}=10.5gc{m}^{-3}$