Q37 / 150 · MHT CET 2026 (17 April Shift 2) · Chemistry
Calculate the number of atoms in 1 g metal that forms bcc crystal structure $[\rho \times a^3 = 6 \cdot 6 \times 10^{-22} \text{ g}]$
- A$4 \cdot 05 \times 10^{21}$
- B$5 \cdot 12 \times 10^{21}$
- Correct answerC$3 \cdot 03 \times 10^{21}$
- D$2 \cdot 47 \times 10^{21}$
Solution
The mass of one unit cell is given by the product of its density and volume:
Mass of unit cell $= \rho \times a^3 = 6 \cdot 6 \times 10^{-22} \text{ g}$
For a body-centered cubic (bcc) crystal structure, the number of atoms per unit cell is $Z = 2$.
Mass of one atom $= \dfrac{\text{Mass of unit cell}}{Z} = \dfrac{6 \cdot 6 \times 10^{-22}}{2} = 3 \cdot 3 \times 10^{-22} \text{ g}$
The number of atoms in $1 \text{ g}$ of the metal is:
Number of atoms $= \dfrac{1 \text{ g}}{\text{Mass of one atom}} = \dfrac{1}{3 \cdot 3 \times 10^{-22}} = 3 \cdot 03 \times 10^{21}$
Answer: $3 \cdot 03 \times 10^{21}$
Mass of unit cell $= \rho \times a^3 = 6 \cdot 6 \times 10^{-22} \text{ g}$
For a body-centered cubic (bcc) crystal structure, the number of atoms per unit cell is $Z = 2$.
Mass of one atom $= \dfrac{\text{Mass of unit cell}}{Z} = \dfrac{6 \cdot 6 \times 10^{-22}}{2} = 3 \cdot 3 \times 10^{-22} \text{ g}$
The number of atoms in $1 \text{ g}$ of the metal is:
Number of atoms $= \dfrac{1 \text{ g}}{\text{Mass of one atom}} = \dfrac{1}{3 \cdot 3 \times 10^{-22}} = 3 \cdot 03 \times 10^{21}$
Answer: $3 \cdot 03 \times 10^{21}$
Hence, the correct answer is option: 'C'.