Iron has body centred cubic cell with a cell edge of 286.5 pm. The density of iron is 7.87 g `cm^(-3)`. Use this information to calculate Avogadro's number. (Atomic mass of Fe = 56 `mol^(-3)`)

Explanation

To calculate Avogadro's number using the given information about iron, we can follow these steps:

### Step 1: Write down the formula for density
The formula for density (D) in terms of the number of atoms in the unit cell (Z), molar mass (M), and the volume of the unit cell (V) is given by:

\[ D = \frac{Z \cdot M}{V \cdot N_A} \]

Where:
- \(D\) = density (g/cm³)
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass (g/mol)
- \(V\) = volume of the unit cell (cm³)
- \(N_A\) = Avogadro's number (atoms/mol)

### Step 2: Identify the values
From the problem:
- Density of iron, \(D = 7.87 \, \text{g/cm}^3\)
- Molar mass of iron, \(M = 56 \, \text{g/mol}\)
- For body-centered cubic (BCC) structure, \(Z = 2\)
- Cell edge length, \(a = 286.5 \, \text{pm} = 286.5 \times 10^{-10} \, \text{cm}\)

### Step 3: Calculate the volume of the unit cell
The volume of the unit cell \(V\) can be calculated using the formula:

\[ V = a^3 \]

Substituting the value of \(a\):

\[ V = (286.5 \times 10^{-10} \, \text{cm})^3 \]

Calculating this:

\[ V = (286.5 \times 10^{-10})^3 = 2.365 \times 10^{-29} \, \text{cm}^3 \]

### Step 4: Rearrange the density formula to solve for Avogadro's number
Rearranging the density formula to solve for \(N_A\):

\[ N_A = \frac{Z \cdot M}{D \cdot V} \]

### Step 5: Substitute the known values into the equation
Substituting the known values into the equation:

\[ N_A = \frac{2 \cdot 56}{7.87 \cdot 2.365 \times 10^{-29}} \]

Calculating the numerator:

\[ 2 \cdot 56 = 112 \]

Calculating the denominator:

\[ 7.87 \cdot 2.365 \times 10^{-29} = 1.860 \times 10^{-28} \]

### Step 6: Calculate Avogadro's number
Now substituting back into the equation:

\[ N_A = \frac{112}{1.860 \times 10^{-28}} \approx 6.008 \times 10^{23} \, \text{mol}^{-1} \]

### Final Answer
Thus, Avogadro's number is approximately:

\[ N_A \approx 6.008 \times 10^{23} \, \text{mol}^{-1} \]