Silver has a cubic unit cell with a cell edge of 408 pm. Its density is 10.6 g `cm^(-3)`. How many atoms of silver are there in the unit cell? What is the structure of silver?

Options

A. 2 , fcc
B. 4, bcc
C. 4, fcc
D. 1, ccp

Answer

To solve the problem, we need to determine how many atoms of silver are present in its unit cell and identify its structure. Here’s a step-by-step solution:

### Step 1: Understand the given data
- **Cell edge length (A)** = 408 pm = 408 × 10^(-12) m
- **Density (D)** = 10.6 g/cm³ = 10.6 × 10^3 kg/m³ (since 1 g/cm³ = 1000 kg/m³)
- **Molar mass (M)** of silver (Ag) = 108 g/mol = 0.108 kg/mol
- **Avogadro's number (NA)** = 6.022 × 10^23 mol^(-1)

### Step 2: Use the density formula
The formula relating density, molar mass, and the number of atoms in the unit cell is given by:

\[ D = \frac{Z \times M}{A^3 \times N_A} \]

Where:
- \( Z \) = number of atoms per unit cell
- \( A \) = edge length of the unit cell
- \( N_A \) = Avogadro's number

### Step 3: Rearranging the formula to find Z
Rearranging the formula to solve for \( Z \):

\[ Z = \frac{D \times A^3 \times N_A}{M} \]

### Step 4: Substitute the values into the equation
Now we substitute the values into the equation:

1. Convert the edge length from pm to m:
\[ A = 408 \times 10^{-12} \text{ m} \]

2. Calculate \( A^3 \):
\[ A^3 = (408 \times 10^{-12})^3 = 6.785 \times 10^{-34} \text{ m}^3 \]

3. Substitute the values:
\[ Z = \frac{(10.6 \times 10^3) \times (6.785 \times 10^{-34}) \times (6.022 \times 10^{23})}{0.108} \]

### Step 5: Calculate Z
Now, performing the calculation step-by-step:

1. Calculate the numerator:
\[ 10.6 \times 10^3 \times 6.785 \times 10^{-34} \times 6.022 \times 10^{23} \approx 4.55 \times 10^{-7} \]

2. Divide by the molar mass:
\[ Z = \frac{4.55 \times 10^{-7}}{0.108} \approx 4.21 \]

### Step 6: Round Z to the nearest whole number
Since \( Z \) must be a whole number, we round 4.21 to 4. Thus, \( Z = 4 \).

### Step 7: Identify the structure of silver
Silver has a face-centered cubic (FCC) structure, which is consistent with having 4 atoms per unit cell.

### Final Answer
- **Number of atoms in the unit cell (Z)**: 4
- **Structure of silver**: Face-Centered Cubic (FCC)