Iron crystallizes in body centered cubic system with edge length `2.86Å`. The density of iron is nearly `X` g/ml. What is the value of `X` here?
Report your answer by rounding it upto nearest whole number.

Answer

To find the density of iron that crystallizes in a body-centered cubic (BCC) system with an edge length of \(2.86 \, \text{Å}\), we can use the formula for density:

\[ \text{Density} = \frac{Z \cdot M}{N_A \cdot A^3} \]

Where:
- \(Z\) = number of atoms per unit cell (for BCC, \(Z = 2\))
- \(M\) = molar mass of iron (approximately \(56 \, \text{g/mol}\))
- \(N_A\) = Avogadro's number (\(6.02 \times 10^{23} \, \text{mol}^{-1}\))
- \(A\) = edge length of the unit cell in cm

### Step 1: Convert the edge length from Ångströms to centimeters
Given that \(1 \, \text{Å} = 10^{-8} \, \text{cm}\):

\[ A = 2.86 \, \text{Å} = 2.86 \times 10^{-8} \, \text{cm} \]

### Step 2: Calculate the volume of the unit cell
The volume \(V\) of the unit cell is given by:

\[ V = A^3 = (2.86 \times 10^{-8} \, \text{cm})^3 \]

Calculating this gives:

\[ V = 2.86^3 \times (10^{-8})^3 = 23.3 \times 10^{-24} \, \text{cm}^3 \]

### Step 3: Substitute the values into the density formula
Now substituting the values into the density formula:

\[ \text{Density} = \frac{Z \cdot M}{N_A \cdot A^3} = \frac{2 \cdot 56}{6.02 \times 10^{23} \cdot 23.3 \times 10^{-24}} \]

### Step 4: Calculate the density
Calculating the numerator:

\[ 2 \cdot 56 = 112 \, \text{g} \]

Now calculating the denominator:

\[ 6.02 \times 10^{23} \cdot 23.3 \times 10^{-24} \approx 1.4 \times 10^{1} \approx 14 \]

Now substituting back into the density formula:

\[ \text{Density} = \frac{112}{14} = 8 \, \text{g/ml} \]

### Step 5: Round off the answer
The density of iron, rounded to the nearest whole number, is:

\[ \text{Density} \approx 8 \, \text{g/ml} \]

### Final Answer
Thus, the value of \(X\) is \(8\).

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