Question

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

• A. $10.5\,g\,cm^{-3}$
• B. $12.5 \,g\, cm^{3}$
• C. $5\,g\,cm^{-2}$
• D. $15.7\,g\, cm^{-3}$

Answer 1

Answer: (A)

Since the lattice is $ccp$ the number of silver atoms per unit cell $= Z = 4$
Molar mass of silver $=107.9\,g\,mol^{-1}=107.9\times 10^{-3}\,kg\,mol^{-1}$
Edge length of unit cell $=a=408.6\,pm = 408.6 \times 10^{-12}\,m$
Density, $d=\frac{Z. M}{a^{3}\cdot N_{A}}$
$=\frac{4\times\left(107.9\times10^{-3} kg \,mol^{-1}\right)}{\left(408.6\times10^{-12} m\right)^{3}\left(6.022\times10^{23}\,mol^{-1}\quad\right)}$
$=10.5\times10^{3}\,kg\,m^{-3}=10.5\,g\,cm^{-3}$