Question

An element has a body-centred cubic $(bcc)$ structure with a cell edge of $288pm$. The density of the element is $7.2g/c{m}^3$. How many atoms are present in $208g$ 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$=(288pm{)}^3={\left(288\times 10^{-12}m\right)}^3$

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

Volume of $208g$ of the element

$=\frac{\mathrm{mass}}{\text{ density }}=\frac{208g}{7.2gc{m}^{-3}}=28.88c{m}^3$

Number of unit cells in this volume

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

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

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