The number of atoms in 2.4 g of body centred cubic crystal with edge length 200 pm is (density `= 10 g cm^(-3), N_A = 6 xx 10^(23)` atoms/mol)

Options

A. `6 xx 10^(22)`
B. `6 xx 10^(20)`
C. `6 xx 10^(23)`
D. `6 xx 10^(19)`

Explanation

To find the number of atoms in 2.4 g of a body-centered cubic (BCC) crystal with an edge length of 200 pm, we can follow these steps:

### Step 1: Determine the number of atoms per unit cell (Z)
In a body-centered cubic (BCC) structure, there are:
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- 1 atom at the body center contributing 1 atom.

Thus, the total number of atoms \( Z \) in a BCC unit cell is:
\[ Z = 8 \times \frac{1}{8} + 1 = 2 \]

### Step 2: Convert edge length from pm to cm
The edge length given is 200 pm (picometers). We need to convert this to centimeters:
\[ 200 \text{ pm} = 200 \times 10^{-12} \text{ m} = 200 \times 10^{-10} \text{ cm} = 2 \times 10^{-8} \text{ cm} \]

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

### Step 4: Use the density to find the molar mass (M)
We know the density \( \rho \) of the crystal is 10 g/cm³. The relationship between density, molar mass, and the number of atoms in the unit cell is given by:
\[ \rho = \frac{Z \times M}{V \times N_A} \]
Where:
- \( \rho \) = density (10 g/cm³)
- \( Z \) = number of atoms per unit cell (2)
- \( M \) = molar mass (g/mol)
- \( V \) = volume of the unit cell (cm³)
- \( N_A \) = Avogadro's number (\( 6 \times 10^{23} \) atoms/mol)

Rearranging the formula to find \( M \):
\[ M = \frac{\rho \times V \times N_A}{Z} \]
Substituting the known values:
\[ M = \frac{10 \text{ g/cm}^3 \times 8 \times 10^{-24} \text{ cm}^3 \times 6 \times 10^{23} \text{ atoms/mol}}{2} \]
Calculating:
\[ M = \frac{10 \times 8 \times 6}{2} = \frac{480}{2} = 240 \text{ g/mol} \]

### Step 5: Calculate the number of moles in 2.4 g
Now we can find the number of moles \( n \) in 2.4 g of the substance:
\[ n = \frac{\text{mass}}{M} = \frac{2.4 \text{ g}}{240 \text{ g/mol}} = 0.01 \text{ mol} \]

### Step 6: Calculate the total number of atoms
Finally, we can find the total number of atoms in 2.4 g:
\[ \text{Number of atoms} = n \times N_A = 0.01 \text{ mol} \times 6 \times 10^{23} \text{ atoms/mol} = 6 \times 10^{21} \text{ atoms} \]

### Final Answer
The number of atoms in 2.4 g of the body-centered cubic crystal is \( 6 \times 10^{21} \) atoms.

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