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

Options

A. `5.783` Å
B. `6.783` Å
C. `7.783` Å
D. `8.783` Å

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 1: Identify the parameters
- **Density (D)** = 3.4 g/cm³
- **Molar mass (M)** of CuCl = Atomic mass of Cu (63.5 g/mol) + Atomic mass of Cl (35.5 g/mol) = 63.5 + 35.5 = 99 g/mol
- **Number of atoms in unit cell (Z)** for FCC = 4
- **Avogadro's number (Nₐ)** = 6.022 × 10²³ mol⁻¹

### Step 2: Use the density formula
The formula for density in terms of the unit cell parameters is given by:

\[ D = \frac{Z \cdot M}{A^3 \cdot N_a} \]

Where:
- \(D\) = density
- \(Z\) = number of atoms in the unit cell
- \(M\) = molar mass
- \(A\) = length of the unit cell (edge length)
- \(N_a\) = Avogadro's number

### Step 3: Rearrange the formula to solve for A³
Rearranging the formula to solve for \(A^3\):

\[ A^3 = \frac{Z \cdot M}{D \cdot N_a} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:

\[ A^3 = \frac{4 \cdot 99 \, \text{g/mol}}{3.4 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 5: Calculate A³
Calculating the numerator:

\[ 4 \cdot 99 = 396 \, \text{g/mol} \]

Calculating the denominator:

\[ 3.4 \cdot 6.022 \times 10^{23} \approx 2.04748 \times 10^{24} \, \text{g/cm}^3 \cdot \text{mol}^{-1} \]

Now substituting back:

\[ A^3 = \frac{396}{2.04748 \times 10^{24}} \approx 1.934 \times 10^{-22} \, \text{cm}^3 \]

### Step 6: Calculate A
Taking the cube root to find \(A\):

\[ A = (1.934 \times 10^{-22})^{1/3} \approx 5.783 \times 10^{-8} \, \text{cm} \]

### Step 7: Convert to Angstroms
Since \(1 \, \text{cm} = 10^{10} \, \text{Å}\):

\[ A \approx 5.783 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{Å/cm} \approx 5.783 \, \text{Å} \]

### Final Answer
The length of the unit cell \(A\) is approximately **5.783 Å**.

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