Copper crystallises into a fee lattice. Its edge length is `3.62 xx 10^(-8)` cm. Calculate the density of copper (atomic mass of Cu=63-5 u, `N_A= 6-022 xx 10^(23) mol^(-1)`).

Explanation

To calculate the density of copper (Cu) that crystallizes in a face-centered cubic (FCC) lattice, we can follow these steps:

### Step 1: Identify the parameters
- **Atomic mass of copper (m)** = 63.5 u (grams per mole)
- **Avogadro's number (N_A)** = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)
- **Edge length (a)** = \(3.62 \times 10^{-8} \, \text{cm}\)

### Step 2: Determine the number of atoms per unit cell (z)
For a face-centered cubic (FCC) lattice, the number of atoms per unit cell (z) is 4.

### Step 3: Calculate the volume of the cubic lattice (V)
The volume of the cubic lattice can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of edge length:
\[ V = (3.62 \times 10^{-8} \, \text{cm})^3 \]

### Step 4: Calculate the volume
Calculating the volume:
\[ V = (3.62 \times 10^{-8})^3 = 4.73 \times 10^{-24} \, \text{cm}^3 \]

### Step 5: Use the density formula
The density (\( \rho \)) can be calculated using the formula:
\[ \rho = \frac{z \times m}{V \times N_A} \]
Substituting the values:
\[ \rho = \frac{4 \times 63.5 \, \text{g/mol}}{4.73 \times 10^{-24} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 6: Calculate the density
Calculating the numerator:
\[ 4 \times 63.5 = 254 \, \text{g/mol} \]
Calculating the denominator:
\[ 4.73 \times 10^{-24} \times 6.022 \times 10^{23} = 2.85 \times 10^{-1} \, \text{cm}^3 \]
Now substituting back into the density formula:
\[ \rho = \frac{254 \, \text{g/mol}}{2.85 \times 10^{-1} \, \text{cm}^3} \approx 8.9 \, \text{g/cm}^3 \]

### Final Answer
The density of copper is approximately \(8.9 \, \text{g/cm}^3\).

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