Question

a metal crystallizes in the face - centered cubic (fcc) lattice. the density of the metal is 12020 kg/m³, and the length of a unit cell edge, a, is 389.08 pm. calculate the mass of one metal atom.
mass:
g
identify the metal.
○ gold
○ palladium
○ platinum
○ silver

Explanation:

Step1: Convert edge length to meters

Convert $a = 389.08\ \text{pm}$ to meters:
$389.08\ \text{pm} = 389.08 \times 10^{-12}\ \text{m} = 3.8908 \times 10^{-10}\ \text{m}$

Step2: Calculate unit cell volume

Volume $V$ of FCC unit cell is $a^3$:
$V = (3.8908 \times 10^{-10}\ \text{m})^3 = 5.887 \times 10^{-29}\ \text{m}^3$

Step3: Find mass of unit cell

Use density $
ho = \frac{m}{V}$, solve for $m$:
$m =
ho \times V = 12020\ \text{kg/m}^3 \times 5.887 \times 10^{-29}\ \text{m}^3 = 7.076 \times 10^{-25}\ \text{kg}$
Convert to grams: $7.076 \times 10^{-22}\ \text{g}$

Step4: Mass of one metal atom

FCC has 4 atoms per unit cell. Divide unit cell mass by 4:
$\text{Mass of 1 atom} = \frac{7.076 \times 10^{-22}\ \text{g}}{4} = 1.769 \times 10^{-22}\ \text{g}$

Step5: Molar mass to identify metal

Molar mass $M = \text{Mass of 1 atom} \times N_A$, where $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$:
$M = 1.769 \times 10^{-22}\ \text{g} \times 6.022 \times 10^{23}\ \text{mol}^{-1} \approx 106.5\ \text{g/mol}$

Answer:

mass: $1.77 \times 10^{-22}\ \text{g}$ (or $1.77 \times 10^{-25}\ \text{kg}$)
Identify the metal: D. silver