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.

Answer 2

The density of a crystal follows from the unit cell formula \( \rho = \dfrac{ZM}{N_A a^3} \). Let's check what value of \( Z \) (atoms per unit cell) and unit conversion each option implies.

1. 19.32 g: For a face-centered cubic (FCC) lattice, \( Z = 4 \) (8 corner atoms each contributing \( \tfrac{1}{8} \), plus 6 face-centered atoms each contributing \( \tfrac{1}{2} \), giving \( 8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4 \)). With \( a = 4.07 \times 10^{-8} \) cm, \( a^3 \approx 6.74 \times 10^{-23} \ \text{cm}^3 \), so \( \rho = \dfrac{4 \times 197}{6.022 \times 10^{23} \times 6.74 \times 10^{-23}} \approx 19.3 \ \text{g/cm}^3 \).
2. 20.32 g: This would need either a larger \( Z \) or a smaller cell volume than the actual FCC geometry and given cell length allow.
3. 21.3 g: Similarly too high for the correct \( Z = 4 \) and the given \( a \); would require underestimating \( a^3 \) noticeably.
4. 18.3 g: Slightly too low; would follow from a small overestimate of \( a^3 \) or an error in the value of \( N_A \) used.

Using the correct FCC value of \( Z = 4 \) with the given lattice parameter reproduces approximately 19.3 g/cm\(^3\).

Therefore, the correct answer is 19.32 g.