The number of atoms in 100 g of an FCC crystal with density `d = 10 g//cm^(3)` and cell edge equal to 100 pm, is equal to

Options

A. `2xx10^(25)`
B. `1xx10^(25)`
C. `4xx10^(25)`
D. `3xx10^(25)`

Answer

To find the number of atoms in 100 g of an FCC (Face-Centered Cubic) crystal with a given density and cell edge length, we can follow these steps:

### Step 1: Identify the given data
- Mass of the FCC crystal (W) = 100 g
- Density (d) = 10 g/cm³
- Cell edge length (a) = 100 pm = 100 × 10^(-12) m = 10^(-8) cm

### Step 2: Calculate the volume of the unit cell
The volume (V) of the unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (10^{-8} \, \text{cm})^3 = 10^{-24} \, \text{cm}^3 \]

### Step 3: Use the density formula to find the molar mass (M)
The density formula for a crystal is given by:
\[ d = \frac{Z \cdot M}{V \cdot N_A} \]
Where:
- Z = number of atoms per unit cell (for FCC, Z = 4)
- M = molar mass (g/mol)
- V = volume of the unit cell (cm³)
- \( N_A \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \) mol⁻¹)

Rearranging the formula to find M:
\[ M = \frac{d \cdot V \cdot N_A}{Z} \]
Substituting the known values:
\[ M = \frac{10 \, \text{g/cm}^3 \cdot 10^{-24} \, \text{cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{4} \]

### Step 4: Calculate the molar mass (M)
Calculating M:
\[ M = \frac{10 \cdot 10^{-24} \cdot 6.022 \times 10^{23}}{4} \]
\[ M = \frac{10 \cdot 6.022 \times 10^{-1}}{4} \]
\[ M = \frac{60.22 \times 10^{-1}}{4} = 15.055 \, \text{g/mol} \]

### Step 5: Calculate the number of moles in 100 g
Using the molar mass calculated:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, \text{g}}{15.055 \, \text{g/mol}} \approx 6.64 \, \text{mol} \]

### Step 6: Calculate the total number of atoms
To find the total number of atoms, multiply the number of moles by Avogadro's number:
\[ \text{Total number of atoms} = \text{Number of moles} \times N_A \]
\[ \text{Total number of atoms} = 6.64 \, \text{mol} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \]
\[ \text{Total number of atoms} \approx 4.00 \times 10^{24} \]

### Final Answer
The number of atoms in 100 g of the FCC crystal is approximately \( 4 \times 10^{24} \).

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