What is the volume of a face centred cubic unit cell, when its density is `2.0 g cm^(-3)` and the molar mass of the substance is 60.23 g `mol^(-1)` ?

Options

A. `4xx10^(-22)cm^(3)`
B. `2xx10^(-22)cm^(3)`
C. `44xx10^(-22)cm^(3)`
D. `22xx10^(-22)cm^(3)`

Answer

To find the volume of a face-centered cubic (FCC) unit cell given its density and molar mass, we can use the formula for density:

\[ \text{Density} (d) = \frac{Z \cdot M}{V \cdot N_A} \]

Where:
- \( Z \) = Number of atoms per unit cell
- \( M \) = Molar mass of the substance
- \( V \) = Volume of the unit cell
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \))

### Step 1: Identify the values
- Given density \( d = 2.0 \, \text{g/cm}^3 \)
- Given molar mass \( M = 60.23 \, \text{g/mol} \)
- For FCC, \( Z = 4 \) (since there are 4 atoms per unit cell in a face-centered cubic structure)
- Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

### Step 2: Rearrange the density formula to find the volume
We can rearrange the formula to solve for the volume \( V \):

\[ V = \frac{Z \cdot M}{d \cdot N_A} \]

### Step 3: Substitute the values into the formula
Now, substituting the known values into the equation:

\[ V = \frac{4 \cdot 60.23 \, \text{g/mol}}{2.0 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 4: Calculate the numerator
Calculating the numerator:

\[ 4 \cdot 60.23 = 240.92 \, \text{g/mol} \]

### Step 5: Calculate the denominator
Calculating the denominator:

\[ 2.0 \cdot 6.022 \times 10^{23} = 1.2044 \times 10^{24} \, \text{g/cm}^3 \cdot \text{mol}^{-1} \]

### Step 6: Complete the calculation
Now, substituting back into the volume equation:

\[ V = \frac{240.92}{1.2044 \times 10^{24}} \approx 2.00 \times 10^{-22} \, \text{cm}^3 \]

### Step 7: Final answer
Thus, the volume of the face-centered cubic unit cell is:

\[ V \approx 2.00 \times 10^{-22} \, \text{cm}^3 \]