An element of atomic mass 52 occurs in bcc structure with cell edge length of 288 pm. Calculate the Avodadro's number if density is `7.2gm//cm^(3)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Atomic mass (m) = 52 g/mol
- Edge length (a) = 288 pm = 288 x 10^(-10) cm
- Density (ρ) = 7.2 g/cm³
- For BCC structure, the number of atoms per unit cell (z) = 2

### Step 2: Write the formula for density
The formula for density (ρ) in terms of the number of atoms per unit cell (z), molar mass (m), edge length (a), and Avogadro's number (Nₐ) is given by:
\[ \rho = \frac{z \cdot m}{a^3 \cdot N_a} \]

### Step 3: Rearrange the formula to solve for Avogadro's number (Nₐ)
Rearranging the formula to find Nₐ gives:
\[ N_a = \frac{z \cdot m}{\rho \cdot a^3} \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
- z = 2
- m = 52 g/mol
- ρ = 7.2 g/cm³
- a = 288 x 10^(-10) cm

Calculating \( a^3 \):
\[ a^3 = (288 \times 10^{-10})^3 = 2.38 \times 10^{-29} \text{ cm}^3 \]

Now substituting these values into the equation for Nₐ:
\[ N_a = \frac{2 \cdot 52}{7.2 \cdot 2.38 \times 10^{-29}} \]

### Step 5: Calculate Nₐ
Calculating the numerator:
\[ 2 \cdot 52 = 104 \]
Calculating the denominator:
\[ 7.2 \cdot 2.38 \times 10^{-29} \approx 1.7136 \times 10^{-28} \]
Now substituting these values:
\[ N_a = \frac{104}{1.7136 \times 10^{-28}} \approx 6.07 \times 10^{23} \text{ mol}^{-1} \]

### Final Answer
Thus, the value of Avogadro's number (Nₐ) is approximately:
\[ N_a \approx 6.07 \times 10^{23} \text{ mol}^{-1} \]