Q32 / 150 · MHT CET 2026 (18 April Shift 1) · Chemistry
Calculate the void volume in fcc unit cell if total volume of a unit cell is $6.4 \times 10^{-23}\ \text{cm}^3$ ?
- A$3.172 \times 10^{-23}\ \text{cm}^3$
- B$3.911 \times 10^{-23}\ \text{cm}^3$
- Correct answerC$1.664 \times 10^{-23}\ \text{cm}^3$
- D$2.051 \times 10^{-23}\ \text{cm}^3$
Solution
The packing efficiency of a face-centered cubic (fcc) unit cell is $74\%$.
The percentage of void space in the fcc unit cell is $100\% - 74\% = 26\%$.
Given the total volume of the unit cell is $V = 6.4 \times 10^{-23}\ \text{cm}^3$.
Void volume $= 0.26 \times V$
Void volume $= 0.26 \times 6.4 \times 10^{-23}\ \text{cm}^3 = 1.664 \times 10^{-23}\ \text{cm}^3$
Answer: $1.664 \times 10^{-23}\ \text{cm}^3$
The percentage of void space in the fcc unit cell is $100\% - 74\% = 26\%$.
Given the total volume of the unit cell is $V = 6.4 \times 10^{-23}\ \text{cm}^3$.
Void volume $= 0.26 \times V$
Void volume $= 0.26 \times 6.4 \times 10^{-23}\ \text{cm}^3 = 1.664 \times 10^{-23}\ \text{cm}^3$
Answer: $1.664 \times 10^{-23}\ \text{cm}^3$
Hence, the correct answer is option: 'C'.