A compound `CuCl` has face `-` centred cubic structure. Its density is `3.4g cm^(-3)`. What is the length of unit cell ?

Answer

To find the length of the unit cell for the compound CuCl with a face-centered cubic (FCC) structure and a given density of 3.4 g/cm³, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the formula for density**:
The density (d) of a crystal can be expressed using the formula:
\[ d = \frac{n \cdot M}{N_A \cdot a^3} \]
where:
- \(d\) = density of the compound (g/cm³)
- \(n\) = number of effective atoms per unit cell
- \(M\) = molar mass of the compound (g/mol)
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23}\) mol⁻¹)
- \(a\) = edge length of the unit cell (cm)

2. **Determine the effective number of atoms (n)**:
In a face-centered cubic structure, there are 4 effective atoms per unit cell. This is because:
- There are 8 corner atoms, each contributing 1/8 (total contribution = 1 atom)
- There are 6 face-centered atoms, each contributing 1/2 (total contribution = 3 atoms)
- Thus, total effective atoms \(n = 1 + 3 = 4\).

3. **Calculate the molar mass (M) of CuCl**:
- Molar mass of Cu = 63.5 g/mol
- Molar mass of Cl = 35.5 g/mol
- Therefore, molar mass of CuCl = \(63.5 + 35.5 = 99.0\) g/mol.

4. **Substitute values into the density formula**:
Now, substituting the known values into the density formula:
\[ 3.4 = \frac{4 \cdot 99.0}{6.022 \times 10^{23} \cdot a^3} \]

5. **Rearrange the equation to solve for \(a^3\)**:
Rearranging gives:
\[ a^3 = \frac{4 \cdot 99.0}{3.4 \cdot 6.022 \times 10^{23}} \]

6. **Calculate \(a^3\)**:
- Calculate the numerator: \(4 \cdot 99.0 = 396.0\)
- Calculate the denominator: \(3.4 \cdot 6.022 \times 10^{23} \approx 2.046 \times 10^{24}\)
- Thus,
\[ a^3 = \frac{396.0}{2.046 \times 10^{24}} \approx 1.937 \times 10^{-22} \text{ cm}^3 \]

7. **Find the edge length \(a\)**:
Taking the cube root:
\[ a \approx (1.937 \times 10^{-22})^{1/3} \approx 5.783 \times 10^{-8} \text{ cm} \]

### Final Answer:
The length of the unit cell \(a\) is approximately \(5.783 \times 10^{-8}\) cm or \(5.783\) Å.