Question:
An element with molar mass $2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}$ forms a cubic unit cell with edge length $405 \mathrm{pm}$. If its density is $2.7$ $\times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, the radius of the element is approximately____________ . $\times 10^{-12} \mathrm{~m}$ (to the nearest integer).
Solution:
(143)
$d=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{N}_{\mathrm{A}} \times \text { Volume }}$
$2.7=\frac{\mathrm{Z} \times 27}{6.02 \times 10^{23} \times\left[4.05 \times 10^{-8}\right]^{3}}$
$Z=4 \Rightarrow f c c$ unit cell
For $f c c$ unit cell $4 r=\sqrt{2} a$
$r=\frac{1.414 \times 405}{4}=143.1675 \mathrm{pm}=143.17 \mathrm{pm}$