Question

The density and edge length values for a crystalline element with fcc lattice are 10 g $c{m}^{-3}$ 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 $5\times {10}^{22}$

$\text{Volume of the unit cell}=a^3=(400pm{)}^3$
$=(400\times {10}^{-10}cm{)}^3=64\times {10}^{-24}c{m}^3$
Density of the unit cell $=10g/c{m}^3$
Mass of unit cell = Volume $\times$ density
$=64\times {10}^{-24}\times 10=640\times {10}^{-24}g$
Now, $6.40\times {10}^{-22}g=$ 1 unit cell
$\therefore 32g=\frac{1}{640\times {10}^{-24}}\times 32=0.05\times {10}^{24}=5\times {10}^{22}$
Thus option (b) is correct.