Question

An element with molar mass $2.7 \times 10^{-2}\,kg\,mol^{-1}$ forms a cubic unit cell with edge length $405\, pm$. If its density is $2.7 \times 10^{3}\, kg\,m^{-3}$, number of atoms present per unit cell is

• A. $2$
• B. $4$
• C. $6$
• D. $3$

Answer 1

Answer: (B)

$M$ (molar mass of the element) $=2.7 \times 10^{-2}\,kg\,mol^{-1}$
a (edge length) $=405\,pm=405 \times 10^{-12}m =4.05 \times 10^{-10}\,m$
d (density) $=2.7 \times 10^{3}\,kg\,m^{-3}$
$N_{A}$ (Avogadro’s number) $=6.022 \times 10^{23}\, mol^{-1}$
Using formula, Density $d=\frac{Z \times M}{a^{3}\times N_{A}}$ or, $Z=\frac{d \times a^{3}\times N_{A}}{M}$
or, $Z=\frac{\left(2.7 \times 10^{3}\right)\left(4.05\times 10^{-10}\right)^{3}\left(6.022\times10^{23}\right)}{2.7\times10^{-2}}=4$
Number of atoms of the element present per unit cell $= 4$