Q72 / 180 · AIIMS 2002 · Chemistry
An element (atomic mass $100 \mathrm{~g} / \mathrm{mol}$ ) having BCC structure has unit cell edge $400 \mathrm{pm}$. The density of element is (No. of atom in BCC $(Z)=2$ ).
- A$2.144 \mathrm{~g} / \mathrm{cm}^3$
- B$7.289 \mathrm{~g} / \mathrm{cm}^3$
- Correct answerC$5.188 \mathrm{~g} / \mathrm{cm}^3$
- D$10.376 \mathrm{~g} / \mathrm{cm}^3$.
Solution
Atomic mass of element $=100 \mathrm{~g} / \mathrm{mol}$
Cell edge $=400 \mathrm{pm}=400 \times 10^{-12}=4 \times 10^{-10}$
and number of atoms in $\mathrm{BCC}(Z)=2$.
As the atomic mass of the metal is $100 \mathrm{~g} / \mathrm{mol}$, therefore
mass of each atom $(\mathrm{m})=\frac{100}{6.023 \times 10^{23}}=16.6 \times 10^{-23} \mathrm{~g}$
The volume of unit cell $=\left(4 \times 10^{-8}\right)^3=64 \times 10^{-24} \mathrm{~cm}^3$.
And mass of unit cell $=Z \times m$
$=2 \times\left(16.6 \times 10^{-23}\right)=33.2 \times 10^{-23} \mathrm{~g}$.
Therefore density of element $=\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}$
$$
=\frac{33.2 \times 10^{-23}}{64 \times 10^{-24}} \approx 5.188 \mathrm{~g} / \mathrm{cm}^3 \text {. }
$$
Hence, the correct answer is option: 'C'.