Question

An element has a body-centred cubic $(b c c)$ structure with a cell edge of $288 \,pm$. The density of the element is $7.2 \,g /\, cm ^{3}$. How many atoms are present in $208 \,g$ of the element?

• A. $6.02 \times 10^{24}$ atoms
• B. $12.09 \times 10^{23}$ atoms
• C. $24.16 \times 10^{23}$ atoms
• D. $29.88 \times 10^{24}$ atoms

Answer 1

Answer: (C)

Volume of the unit cell$=(288 \,pm )^{3} =\left(288 \times 10^{-12} m \right)^{3}$

$=\left(288 \times 10^{-10} \,cm \right)^{3} =2.39 \times 10^{-23} \,cm ^{3}$

Volume of $208 \,g$ of the element

$=\frac{\operatorname{mass}}{\text { density }}=\frac{208 g }{7.2\, g\, cm ^{-3}}=28.88\, cm ^{3}$

Number of unit cells in this volume

$=\frac{28.88 \,cm ^{3}}{2.39 \times 10^{-23} \,cm ^{3} / \text { unit cell }}=12.08 \times 10^{23}$

Since, each $b c c$ cubic unit cell contains $2$ atoms, therefore, the total number of atoms in $208 \, g$.

$=2 \text { (atoms/unit cell) } \times 12.08 \times 10^{23} \text { unit cells }=24.16 \times 10^{23} \text { atoms }$