Q12 / 150 · MHT CET 2026 (16 April Shift 2) · Chemistry
When silver crystallizes, it forms face centered cubic cells, if the volume of unit cells $6.84 \times 10^{-23}\ \text{cm}^3$. Calculate the density of silver. (Molar mass of silver is 108 g/mol, $N_A = 6.022 \times 10^{23}$)
- A$12.49\ \text{gm cm}^{-3}$
- Correct answerB$10.49\ \text{gm cm}^{-3}$
- C$16.89\ \text{gm cm}^{-3}$
- D$20.49\ \text{gm cm}^{-3}$
Solution
For a face-centered cubic (fcc) unit cell, the number of atoms per unit cell is $Z = 4$.
The density of the unit cell is given by the formula:
$d = \dfrac{Z \times M}{N_A \times V}$
Substituting the given values:
$d = \dfrac{4 \times 108}{6.022 \times 10^{23} \times 6.84 \times 10^{-23}}$
$d = \dfrac{432}{41.19}$
$d = 10.49\ \text{gm cm}^{-3}$
Answer: $10.49\ \text{gm cm}^{-3}$
The density of the unit cell is given by the formula:
$d = \dfrac{Z \times M}{N_A \times V}$
Substituting the given values:
$d = \dfrac{4 \times 108}{6.022 \times 10^{23} \times 6.84 \times 10^{-23}}$
$d = \dfrac{432}{41.19}$
$d = 10.49\ \text{gm cm}^{-3}$
Answer: $10.49\ \text{gm cm}^{-3}$
Hence, the correct answer is option: 'B'.