Question

An element crystallizes in face-centered cubic lattice with density as $5{\text{g/cm}}^3$ and edge length of the side of unit cell is $400\text{pm}$.
Take $N_A=6\times {10}^{23}$. Then the molar mass of the element in ${\text{g mol}}^{-1}$ is :

Answer 1

$Z=4$ in an FCC lattice

$\rho =5{\text{g/cm}}^3$
$a=400\text{pm}=4\times {10}^{-8}\text{cm}$
$\therefore a^3=64\times {10}^{-24}{\text{cm}}^3$
$N_A=6\times {10}^{23}$
$\rho =\frac{Z\times M}{N_A\times a^3}$

$\therefore M=\frac{\rho N_A{a}^3}{Z}$
$\Rightarrow M=\frac{5{\text{g/cm}}^3\times 6\times {10}^{23}{\text{mol}}^{-1}\times 64\times {10}^{-24}{\text{cm}}^3}{4}$

$\Rightarrow M=48{\text{g mol}}^{-1}$