The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one copper atom at each lattice point. The calculated density of copper in kg/m³ is ________ . [Molar mass of Cu: 63.54 g; Avogadro Number=6.022$\times$10²³]
Show Hint
When calculating crystal density, be extremely careful with units. It's best to convert all quantities to a consistent system (like SI units: kg, m) before plugging them into the formula. Remember Z=1 for simple cubic, Z=2 for BCC, and Z=4 for FCC.
Step 1: Identify the crystal structure
Copper crystallizes in a face-centered cubic (FCC) lattice.
For an FCC lattice:
\[
Z = 4 \quad \text{(number of atoms per unit cell)}
\]
Step 2: Write the density formula
The density of a crystalline solid is given by:
\[
\rho = \frac{Z \times M}{N_A \times a^3}
\]
where
$Z$ = number of atoms per unit cell
$M$ = molar mass
$N_A$ = Avogadro number
$a$ = edge length of the unit cell
Step 3: Convert all quantities to SI units
\[
M = 63.54\ \text{g mol}^{-1} = 63.54 \times 10^{-3}\ \text{kg mol}^{-1}
\]
\[
a = 3.596\ \text{Å} = 3.596 \times 10^{-10}\ \text{m}
\]
\[
N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}
\]
Step 4: Calculate the volume of the unit cell
\[
a^3 = (3.596 \times 10^{-10})^3
\]
\[
a^3 \approx 4.65 \times 10^{-29}\ \text{m}^3
\]
Step 5: Substitute values into the density formula
\[
\rho = \frac{4 \times (63.54 \times 10^{-3})}
{6.022 \times 10^{23} \times 4.65 \times 10^{-29}}
\]
\[
\rho = \frac{0.25416}{2.800 \times 10^{-5}}
\]
\[
\rho \approx 9.08 \times 10^{3}\ \text{kg m}^{-3}
\]
Step 6: Final Answer
\[
\boxed{\rho \approx 9077\ \text{kg m}^{-3}}
\]
