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 follow these steps:

### Step-by-Step Solution:

1. **Identify the formula for density**:
The formula for density (\(d\)) is given by:
\[ d = \frac{Z \cdot M}{N_A \cdot V} \]
where:
- \(Z\) = number of atoms per unit cell (for FCC, \(Z = 4\))
- \(M\) = molar mass of the substance
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \(V\) = volume of the unit cell

2. **Rearrange the formula to find the volume**:
We can rearrange the formula to solve for the volume \(V\):
\[ V = \frac{Z \cdot M}{d \cdot N_A} \]

3. **Substitute the known values**:
Given:
- Density \(d = 2.0 \, \text{g/cm}^3\)
- Molar mass \(M = 60.23 \, \text{g/mol}\)
- Number of atoms in FCC \(Z = 4\)
- Avogadro's number \(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)

Now substitute these values into the rearranged formula:
\[ V = \frac{4 \cdot 60.23}{2.0 \cdot 6.022 \times 10^{23}} \]

4. **Calculate the numerator and denominator**:
- Numerator: \(4 \cdot 60.23 = 240.92\)
- Denominator: \(2.0 \cdot 6.022 \times 10^{23} = 12.044 \times 10^{23}\)

5. **Perform the division**:
\[ V = \frac{240.92}{12.044 \times 10^{23}} = 20.00 \times 10^{-23} \, \text{cm}^3 \]

6. **Express in scientific notation**:
\[ V = 2.00 \times 10^{-22} \, \text{cm}^3 \]

### Final Answer:
The volume of the face-centered cubic unit cell is:
\[ V = 2.00 \times 10^{-22} \, \text{cm}^3 \]