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: For FCC gold, \( Z = 4 \) atoms per unit cell, \( M = 197 \) g/mol, \( a = 4.07 \times 10^{-8} \) cm, so \( a^3 \approx 6.74 \times 10^{-23} \ \text{cm}^3 \).
Step 2: Apply \( \rho = \dfrac{ZM}{N_A a^3} = \dfrac{4 \times 197}{6.022 \times 10^{23} \times 6.74 \times 10^{-23}} \).
Step 3: Evaluating the fraction.
\[ \boxed{\rho \approx 19.32 \ \text{g/cm}^3} \]

Answer 2

Instead of computing the density directly, let's reason about the effect of each factor on the result and check the options against that reasoning.

1. 19.32 g: Using the correct FCC packing factor \( Z = 4 \), Avogadro's number, and the cubed lattice length in centimeters, the mass-per-unit-cell divided by the volume-per-unit-cell gives a density in this range, matching known experimental gold density values closely.
2. 20.32 g: Overestimates density; would require the cell edge to be effectively shorter than 4.07 Å, which isn't the case here.
3. 21.3 g: Even higher, and would require both a shorter effective cell edge and a higher atom count per cell than FCC actually has.
4. 18.3 g: Underestimates density; would follow if the cell edge used were slightly larger than 4.07 Å or if \( Z \) were taken as less than 4.

With the correct FCC atom count and lattice parameter, the density comes out closest to 19.32 g/cm\(^3\), which also matches gold's well-documented real density.

Therefore, the correct answer is 19.32 g.