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

Options

A. `4xx10^25`
B. `2xx10^25`
C. `8xx10^25`
D. `1xx10^25`

Answer

To find the number of atoms in a 100 g FCC crystal with a density of 10 g/cm³ and an edge length of 100 pm, we can follow these steps:

### Step 1: Understand the FCC structure
In a face-centered cubic (FCC) structure, there are 4 atoms per unit cell. This is denoted by \( Z = 4 \).

### Step 2: Calculate the volume of the unit cell
The edge length \( a \) is given as 100 pm (picometers), which we need to convert to centimeters for consistency with the density units:
\[ a = 100 \, \text{pm} = 100 \times 10^{-12} \, \text{m} = 100 \times 10^{-10} \, \text{cm} = 1 \times 10^{-8} \, \text{cm} \]

Now, calculate the volume \( V \) of the unit cell:
\[ V = a^3 = (1 \times 10^{-8} \, \text{cm})^3 = 1 \times 10^{-24} \, \text{cm}^3 \]

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

Rearranging the formula to find \( M \):
\[ M = \frac{d \times V \times N_A}{Z} \]

Substituting the known values:
\[ M = \frac{10 \, \text{g/cm}^3 \times 1 \times 10^{-24} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{4} \]
\[ M = \frac{10 \times 1 \times 6.022}{4} \times 10^{-1} \, \text{g/mol} \]
\[ M = \frac{60.22}{4} \times 10^{-1} \, \text{g/mol} = 15.055 \, \text{g/mol} \]

### Step 4: Calculate the number of moles in 100 g
To find the number of moles in 100 g of the substance:
\[ \text{Number of moles} = \frac{\text{mass}}{M} = \frac{100 \, \text{g}}{15.055 \, \text{g/mol}} \approx 6.64 \, \text{mol} \]

### Step 5: Calculate the number of atoms
Now, we can calculate the total number of atoms using Avogadro's number:
\[ \text{Number of atoms} = \text{Number of moles} \times N_A \]
\[ \text{Number of atoms} = 6.64 \, \text{mol} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \]
\[ \text{Number of atoms} \approx 4.00 \times 10^{24} \]

### Final Answer
The number of atoms in 100 g of the FCC crystal is approximately \( 4.00 \times 10^{24} \).

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