Q11 / 150 · MHT CET 2026 (20 April Shift 1) · Chemistry
Calculate the density of an element having molar mass $225\ \text{g mol}^{-1}$ forming bcc structure $[a^3 \times N_A = 75\ \text{cm}^3\text{mol}^{-1}]$
- Correct answerA$6.0$ g cm$^{-3}$
- B$2.81$ g cm$^{-3}$
- C$9.24$ g cm$^{-3}$
- D$11.36$ g cm$^{-3}$
Solution
For a body-centered cubic (bcc) structure, the number of atoms per unit cell is $Z = 2$.
The formula for the density of a unit cell is given by:
$d = \dfrac{Z \times M}{a^3 \times N_A}$
Substituting the given values $Z = 2$, $M = 225\ \text{g mol}^{-1}$, and $a^3 \times N_A = 75\ \text{cm}^3\text{mol}^{-1}$:
$d = \dfrac{2 \times 225}{75}$
$d = \dfrac{450}{75} = 6.0\ \text{g cm}^{-3}$
Answer: $6.0$ g cm$^{-3}$
The formula for the density of a unit cell is given by:
$d = \dfrac{Z \times M}{a^3 \times N_A}$
Substituting the given values $Z = 2$, $M = 225\ \text{g mol}^{-1}$, and $a^3 \times N_A = 75\ \text{cm}^3\text{mol}^{-1}$:
$d = \dfrac{2 \times 225}{75}$
$d = \dfrac{450}{75} = 6.0\ \text{g cm}^{-3}$
Answer: $6.0$ g cm$^{-3}$
Hence, the correct answer is option: 'A'.