Question

Gold crystallizes in a face-centered cubic lattice with unit cell length of 4.07 Å. Relative atomic mass of gold is 197.0. The density of gold will be

• A. 19.32 g
• B. 20.32 g
• C. 21.3 g
• D. 18.3 g

Hint:

For calculating density from the unit cell structure, use the formula involving Avogadro's number and the unit cell volume.

Answer 1

The Correct Option is A

Step 1: Formula for density.
Density (\( \rho \)) is given by the formula: \[ \rho = \frac{Z M}{N_A a^3} \] where: - \( Z \) is the number of atoms per unit cell (for face-centered cubic, \( Z = 4 \)), - \( M \) is the molar mass of gold (197 g/mol), - \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \)), - \( a \) is the unit cell length (4.07 Å = \( 4.07 \times 10^{-10} \) m). Step 2: Substituting the values.
Substitute the values into the density formula: \[ \rho = \frac{4 \times 197}{6.022 \times 10^{23} \times (4.07 \times 10^{-10})^3} \approx 19.32 \, \text{g/cm}^3 \] Step 3: Conclusion.
The correct density of gold is \( \boxed{19.32} \, \text{g/cm}^3 \). The correct answer is (1) 19.32 g.