Question

The structure of a cell of an element is body-centred cubic (bcc). The length of the core of the cell is 200 pm. Density of the element is 7 g/cm\(^3\). Determine the number of atoms in 20 g element.

Hint:

In bcc, 2 atoms per unit cell; always use density formula: \(\rho = \frac{Z \times M}{a^3 \times N_A}\).

Answer 1

Step 1: Relation in bcc structure.
In bcc, body diagonal = \(4r = \sqrt{3}a\).
Here, edge length \(a = 200 \, \text{pm} = 2 \times 10^{-8} \, \text{cm}\).
Step 2: Volume of unit cell.
\[ a^3 = (2 \times 10^{-8})^3 = 8 \times 10^{-24} \, \text{cm}^3 \] Step 3: Mass of unit cell.
\[ \text{Mass} = \text{Density} \times \text{Volume} = 7 \times 8 \times 10^{-24} = 5.6 \times 10^{-23} \, g \] Step 4: Atoms per unit cell.
In bcc: 2 atoms/unit cell.
So, molar mass = \(\frac{\text{Mass of unit cell} \times N_A}{2}\). \[ M = \frac{5.6 \times 10^{-23} \times 6.022 \times 10^{23}}{2} \approx 17 \, g/mol \] Step 5: Number of atoms in 20 g.
\[ \text{Moles} = \frac{20}{17} \approx 1.18 \, mol \] \[ \text{Atoms} = 1.18 \times 6.022 \times 10^{23} \approx 1.2 \times 10^{23} \] Step 6: Conclusion.
Number of atoms in 20 g element = \(\approx 1.2 \times 10^{23}\).