A crystalline solid of pure substance has a face - centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is `"8 g cm"^(-3)`, then the number of atoms present in 128 g of the crystal is

Options

A. `2xx10^(24)`
B. `4xx10^(24)`
C. `1xx10^(24)`
D. `3xx10^(24)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Face-Centred Cubic (FCC) Structure
In a face-centred cubic (FCC) structure, there are 4 atoms per unit cell. This is derived from the fact that there are 8 atoms at the corners (each contributing 1/8 of an atom) and 6 atoms on the faces (each contributing 1/2 of an atom).

**Hint:** Remember that in FCC, the total number of atoms (Z) is calculated as:
\[ Z = \text{(atoms at corners)} + \text{(atoms on faces)} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4 \]

### Step 2: Convert the Cell Edge Length from Picometers to Centimeters
The edge length of the unit cell is given as 400 pm (picometers). We need to convert this to centimeters for consistency with the density units.

\[ 400 \text{ pm} = 400 \times 10^{-12} \text{ m} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm} \]

**Hint:** Remember that \(1 \text{ pm} = 10^{-12} \text{ m}\) and \(1 \text{ cm} = 10^{-2} \text{ m}\).

### Step 3: Calculate the Volume of the Unit Cell
The volume of the cubic unit cell can be calculated using the formula:
\[ \text{Volume} = \text{edge length}^3 = (4 \times 10^{-8} \text{ cm})^3 = 64 \times 10^{-24} \text{ cm}^3 = 6.4 \times 10^{-23} \text{ cm}^3 \]

**Hint:** Use the formula for the volume of a cube, \( V = a^3 \), where \( a \) is the edge length.

### Step 4: Use the Density Formula to Find the Molar Mass
The density (\( \rho \)) is given as 8 g/cm³. The relationship between density, molar mass (M), and the number of atoms per unit cell (Z) is given by:
\[ \rho = \frac{Z \cdot M}{N_A \cdot V} \]
Where:
- \( \rho \) = density
- \( Z \) = number of atoms per unit cell (4 for FCC)
- \( M \) = molar mass
- \( N_A \) = Avogadro's number (\(6.022 \times 10^{23} \text{ mol}^{-1}\))
- \( V \) = volume of the unit cell

Rearranging the formula to find \( M \):
\[ M = \frac{\rho \cdot N_A \cdot V}{Z} \]
Substituting the values:
\[ M = \frac{8 \text{ g/cm}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1} \cdot 6.4 \times 10^{-23} \text{ cm}^3}{4} \]

Calculating \( M \):
\[ M = \frac{8 \cdot 6.022 \cdot 6.4}{4} = 96.768 \text{ g/mol} \]

**Hint:** Make sure to keep track of units when substituting values into the formula.

### Step 5: Calculate the Number of Moles in 128 g of the Substance
Now, we can find the number of moles (\( n \)) in 128 g of the substance using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{128 \text{ g}}{96.768 \text{ g/mol}} \approx 1.32 \text{ mol} \]

**Hint:** The number of moles is calculated using the mass of the substance divided by its molar mass.

### Step 6: Calculate the Number of Atoms
To find the total number of atoms, we multiply the number of moles by Avogadro's number:
\[ \text{Number of atoms} = n \cdot N_A = 1.32 \text{ mol} \cdot 6.022 \times 10^{23} \text{ mol}^{-1} \approx 7.95 \times 10^{23} \text{ atoms} \]

**Hint:** Remember that the total number of atoms is the product of the number of moles and Avogadro's number.

### Final Answer
The number of atoms present in 128 g of the crystal is approximately \( 7.95 \times 10^{23} \) atoms.