The Cr metal (atomic weight 52) crystallises BBC structure. The unit cell edge length is 2.88 Å. The density of Cr is `7.2 g ml^(-1)`. The number of atom is 52 g of Cr is

Options

A. `3 xx 10^(23)`
B. `12 xx 10^(23)`
C. `6 xx 10^(23)`
D. `9 xx 10^(23)`

Answer

To find the number of atoms in 52 g of chromium (Cr), we will follow these steps:

### Step 1: Understand the relationship between density, molar mass, and unit cell parameters
We know that the density (D) of a substance can be expressed as:
\[ D = \frac{Z \cdot M}{A^3 \cdot N_A} \]
Where:
- \( Z \) = number of atoms in the unit cell
- \( M \) = molar mass (g/mol)
- \( A \) = edge length of the unit cell (cm)
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol\(^{-1}\))

### Step 2: Rearranging the formula to find molar mass (M)
From the density formula, we can rearrange it to find the molar mass:
\[ M = \frac{D \cdot A^3 \cdot N_A}{Z} \]

### Step 3: Calculate the volume of the unit cell
Convert the edge length from angstroms to centimeters:
- Given \( A = 2.88 \, \text{Å} = 2.88 \times 10^{-8} \, \text{cm} \)
- Volume of the unit cell \( V = A^3 = (2.88 \times 10^{-8} \, \text{cm})^3 \)

Calculating \( A^3 \):
\[ A^3 = (2.88 \times 10^{-8})^3 = 2.39 \times 10^{-23} \, \text{cm}^3 \]

### Step 4: Substitute values into the molar mass formula
Given:
- Density \( D = 7.2 \, \text{g/cm}^3 \)
- For BCC structure, \( Z = 2 \)

Now substituting into the molar mass formula:
\[ M = \frac{7.2 \, \text{g/cm}^3 \cdot 2.39 \times 10^{-23} \, \text{cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{2} \]

Calculating \( M \):
\[ M = \frac{7.2 \cdot 2.39 \cdot 6.022}{2} \]
\[ M = \frac{7.2 \cdot 14.39}{2} \]
\[ M = \frac{103.61}{2} = 51.805 \, \text{g/mol} \]

### Step 5: Calculate the number of moles in 52 g of Cr
Using the molar mass we calculated:
\[ \text{Number of moles} = \frac{\text{mass}}{M} = \frac{52 \, \text{g}}{51.805 \, \text{g/mol}} \approx 1.004 \, \text{mol} \]

### Step 6: Calculate the number of atoms
Using Avogadro's number:
\[ \text{Number of atoms} = \text{Number of moles} \times N_A \]
\[ \text{Number of atoms} = 1.004 \, \text{mol} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \approx 6.04 \times 10^{23} \]

### Final Answer
The number of atoms in 52 g of Cr is approximately \( 6.04 \times 10^{23} \).

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