The no. of atoms in 100g of a fcc crystal with density =10.0 g/cc and edge length as 100 pm is :

Options

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

Explanation

To find the number of atoms in 100 g of a face-centered cubic (FCC) crystal with a density of 10.0 g/cm³ and an edge length of 100 pm, we can follow these steps:

### Step 1: Understand the formula for density
The density (d) of a crystal can be expressed using the formula:
\[ d = \frac{Z \cdot M_w}{N_A \cdot a^3} \]
where:
- \( Z \) = number of atoms per unit cell (for FCC, \( Z = 4 \))
- \( M_w \) = molecular weight
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol⁻¹)
- \( a \) = edge length of the unit cell

### Step 2: Convert edge length to centimeters
The edge length is given as 100 pm (picometers). We need to convert this to centimeters:
\[ 100 \text{ pm} = 100 \times 10^{-12} \text{ m} = 100 \times 10^{-10} \text{ cm} = 10^{-8} \text{ cm} \]

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

### Step 4: Substitute values into the density formula
Now we can substitute the known values into the density formula:
\[ 10 \text{ g/cm}^3 = \frac{4 \cdot M_w}{6.022 \times 10^{23} \cdot 10^{-24}} \]

### Step 5: Rearranging to find molecular weight
Rearranging the equation to solve for \( M_w \):
\[ M_w = \frac{10 \cdot 6.022 \times 10^{23} \cdot 10^{-24}}{4} \]
\[ M_w = \frac{10 \cdot 6.022}{4} \text{ g/mol} \]
\[ M_w = 15.055 \text{ g/mol} \]

### Step 6: Calculate the number of moles in 100 g
Using the molecular weight, we can find the number of moles in 100 g:
\[ \text{Number of moles} = \frac{\text{mass}}{M_w} = \frac{100 \text{ g}}{15.055 \text{ g/mol}} \]
\[ \text{Number of moles} \approx 6.64 \text{ mol} \]

### Step 7: Calculate the number of atoms
To find the total number of atoms, we multiply the number of moles by Avogadro's number:
\[ \text{Number of atoms} = \text{Number of moles} \times N_A \]
\[ \text{Number of atoms} = 6.64 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \]
\[ \text{Number of atoms} \approx 4.00 \times 10^{24} \text{ atoms} \]

### Final Answer
The number of atoms in 100 g of the FCC crystal is approximately:
\[ 4 \times 10^{25} \text{ atoms} \]