Question

An element (atomic mass $=50\text{g/mol}$) that has a fcc structure has a unit cell edge length of $200\text{pm}$. The density of the element is:
(Take $N_A=6\times {10}^{23}$)

• A. $10{\text{g cm}}^{-3}$
• B. $15.67{\text{g cm}}^{-3}$
• C. $17.42{\text{g cm}}^{-3}$
• D. $41.66{\text{g cm}}^{-3}$

Answer 1

The correct option is D $41.66{\text{g cm}}^{-3}$

Density $\rho$ is given by
$\rho =\frac{ZM}{a^3\times N_A}$
Since it has a FCC structure, $Z=4$

$\therefore \rho =\frac{4\times 50}{(200{)}^3\times 6\times {10}^{23}\times {10}^{-30}}$

$\Rightarrow \rho =41.66{\text{g cm}}^{-3}$
FCC Unit Cell: