Question

An element forms a body centered cubic (bcc) lattice with edge length of $300 \,pm$. If the density of the element is $7.2 \,g\, cm ^{-3}$, the number of atoms present in $324\, g$ of it approximately is

• A. $3.33 \times 10^{23}$
• B. $6.66 \times 10^{23}$
• C. $3.33 \times 10^{24}$
• D. $6.66 \times 10^{24}$

Answer 1

Answer: (C)

Volume of bcc unit cell = $300\,pm$

$=\left(300 \times 10^{-10}\, cm \right)^{3}=27 \times 10^{-24} \,cm ^{3}$

Volume of $324\, g$ of the element $=\frac{\text { Mass }}{\text { Density }}$

$=\frac{324 \,g }{7.2\, g \,cm ^{-3}}=45\, cm ^{3}$

For a bcc structure, number of atoms per unit cell $=2$

So, number of atoms $=\frac{\text { Total volume }}{\text { Volume of a unit cell }}$

$=\frac{2 \times 45}{27} \times 10^{24}=3.33 \times 10^{24}$