Question

The density and edge length values for a crystalline element with $ fcc $ lattice are $ 10 \,g \,cm^{-3} $ and $ 400\, pm $ , respectively. The number of unit cells in $ 32 \,g $ of this crystal is

• A. $ 8\times 10^{23} $
• B. $ 5\times 10^{22} $
• C. $ 8\times 10^{22} $
• D. $ 5\times 10^{23} $

Answer 1

The Correct Option is B

Volume of the unit cell $=a^{3}=(400\,pm)^{3}$
$=64\times 10^{-30}\,m^{3}=64 \times 10^{-24}\,cm^{3}$
Density of the unit cell $=10 \,g/cm^{3}$
Mass of unit cell = Volume $\times$ density
$=64 \times 10^{-24}\times10$
$=640 \times 10^{-24}\,g$
Now, $640 \times 10^{-24}\,g \equiv 1$ unit cell
$\therefore 32\,g \equiv \frac{1}{640 \times 10^{-24}}\times32$
$=0.05\times 10^{24}$
$=5\times 10^{22}$