Q12 / 150 · MHT CET 2026 (15 April Shift 2) · Chemistry
Calculate the number of atoms present in 1.625 g metal if it forms bcc unit cell structure.
$[ \rho \times a^3 = 3.25 \times 10^{-22}\ \text{g}]$
$[ \rho \times a^3 = 3.25 \times 10^{-22}\ \text{g}]$
- A$2.0 \times 10^{22}$
- B$1.5 \times 10^{22}$
- C$2.5 \times 10^{22}$
- Correct answerD$1.0 \times 10^{22}$
Solution
Mass of one unit cell is given by $\rho \times a^3 = 3.25 \times 10^{-22}$ g.
For a body-centered cubic (bcc) structure, the number of atoms per unit cell is $Z = 2$.
Mass of one atom = $\dfrac{\text{Mass of unit cell}}{Z} = \dfrac{3.25 \times 10^{-22}}{2} = 1.625 \times 10^{-22}$ g.
Total number of atoms in $1.625$ g of the metal = $\dfrac{\text{Total mass}}{\text{Mass of one atom}}$
Number of atoms = $\dfrac{1.625}{1.625 \times 10^{-22}} = 1.0 \times 10^{22}$
Answer: $1.0 \times 10^{22}$
For a body-centered cubic (bcc) structure, the number of atoms per unit cell is $Z = 2$.
Mass of one atom = $\dfrac{\text{Mass of unit cell}}{Z} = \dfrac{3.25 \times 10^{-22}}{2} = 1.625 \times 10^{-22}$ g.
Total number of atoms in $1.625$ g of the metal = $\dfrac{\text{Total mass}}{\text{Mass of one atom}}$
Number of atoms = $\dfrac{1.625}{1.625 \times 10^{-22}} = 1.0 \times 10^{22}$
Answer: $1.0 \times 10^{22}$
Hence, the correct answer is option: 'D'.