An fcc lattice has a lattice parameter `a = 400` pm. Calculater the molar volume of the lattice including all the empty space.

Options

A. `10.8 mL`
B. `96 mL`
C. `8.6 mL`
D. `9.6 mL`

Explanation

To calculate the molar volume of an FCC lattice with a lattice parameter \( a = 400 \) pm, we will follow these steps:

### Step 1: Calculate the volume of the unit cell
The volume \( V \) of a cubic unit cell is given by the formula:
\[ V = a^3 \]
where \( a \) is the lattice parameter. First, we need to convert the lattice parameter from picometers to centimeters:
\[ a = 400 \, \text{pm} = 400 \times 10^{-10} \, \text{cm} = 4 \times 10^{-8} \, \text{cm} \]
Now, we can calculate the volume:
\[ V = (4 \times 10^{-8} \, \text{cm})^3 = 64 \times 10^{-24} \, \text{cm}^3 \]

### Step 2: Calculate the volume of \( N_A \) unit cells
To find the total volume of \( N_A \) unit cells (where \( N_A \) is Avogadro's number, approximately \( 6.022 \times 10^{23} \)):
\[ \text{Volume of } N_A \text{ unit cells} = V \times N_A = 64 \times 10^{-24} \, \text{cm}^3 \times 6.022 \times 10^{23} \]
Calculating this gives:
\[ \text{Volume of } N_A \text{ unit cells} = 64 \times 6.022 \times 10^{-1} \, \text{cm}^3 = 38.4 \, \text{cm}^3 \]

### Step 3: Calculate the molar volume
For an FCC lattice, the effective number of atoms per unit cell \( Z \) is 4. The molar volume \( V_m \) can be calculated using the formula:
\[ V_m = \frac{\text{Volume of } N_A \text{ unit cells}}{Z} \]
Substituting the values we have:
\[ V_m = \frac{38.4 \, \text{cm}^3}{4} = 9.6 \, \text{cm}^3 \]

### Final Answer
The molar volume of the FCC lattice including all the empty space is \( 9.6 \, \text{cm}^3 \).

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