Q6 / 155 · WBJEE 2021 · Chemistry
An element crystallizes in a body centred cubic lattice. The edge length of the unit cell is $200 \mathrm{pm}$ and the density of the element is $5.0 \mathrm{~g} \mathrm{~cm}^{-3}$. Calculate the number of atoms in $100 \mathrm{~g}$ of this element.
- A$2.5 \times 10^{23}$
- B$2.5 \times 10^{24}$
- C$5.0 \times 10^{23}$
- Correct answerD$5.0 \times 10^{24}$
Solution
$\rho_{\text {unit cell }}(\mathrm{g} / \mathrm{cc})=\frac{\mathrm{Z} \times \mathrm{M} \cdot \mathrm{Mass}(\mathrm{g} / \mathrm{mol})}{\mathrm{N}_{\mathrm{A}}\left(\mathrm{mol}^{-1}\right) \times(\mathrm{a} \mathrm{pm})^{3} \times 10^{-30}} \quad \mathrm{Z}_{\text {B.c.c }}=2$
or $5=\frac{2 \times \text { M.Mass }}{\mathrm{N}_{\mathrm{A}} \times(200)^{3} \times 10^{-30}}$
or M.Mass $=\frac{5 \times 8 \times \mathrm{N}_{\mathrm{A}} \times 10^{-24}}{2}=2 \mathrm{~N}_{\mathrm{A}} \times 10^{-23}$
Now, no. of moles in $100 \mathrm{~g}=\frac{100}{2 \mathrm{~N}_{\mathrm{A}} \times 10^{-23}}$
$\therefore$ no. of atoms $=\frac{100}{2 \mathrm{~N}_{\mathrm{A}} \times 10^{-23}} \times \mathrm{N}_{\mathrm{A}}=50 \times 10^{23}=5.0 \times 10^{24}$
Hence, the correct answer is option: 'D'.