Question

An element forms CCP lattice. If the length of its core of unit cell is 408.6 pm, then calculate the density of the element. (Atomic weight = 107.9 u)

Hint:

For CCP lattice, there are 4 atoms per unit cell, and the density can be calculated using atomic weight and unit cell volume.

Answer 1

Step 1: Understanding CCP Lattice.
A cubic close-packed (CCP) unit cell contains 4 atoms. The edge length of the unit cell is given as 408.6 pm (1 pm = \( 10^{-12} \) m). Step 2: Volume of Unit Cell.
The volume of the unit cell can be calculated using the formula: \[ \text{Volume of unit cell} = a^3 \] where \( a \) is the edge length of the unit cell. Given that \( a = 408.6 \, \text{pm} = 408.6 \times 10^{-12} \, \text{m} \), we can calculate the volume of the unit cell: \[ V = (408.6 \times 10^{-12})^3 = 6.8 \times 10^{-29} \, \text{m}^3 \] Step 3: Mass of Unit Cell.
The mass of the unit cell is given by the formula: \[ \text{Mass of unit cell} = \frac{\text{Atomic weight}}{\text{Avogadro's number}} \times \text{number of atoms in the unit cell} \] Given that the atomic weight is 107.9 u, Avogadro's number is \( 6.022 \times 10^{23} \, \text{atoms/mol} \), and there are 4 atoms in the unit cell, we get: \[ \text{Mass of unit cell} = \frac{107.9}{6.022 \times 10^{23}} \times 4 = 7.17 \times 10^{-23} \, \text{g} \] Step 4: Calculate the Density.
The density is given by the formula: \[ \text{Density} = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} \] Substituting the values: \[ \text{Density} = \frac{7.17 \times 10^{-23}}{6.8 \times 10^{-29}} = 8.2 \, \text{g/cm}^3 \] Step 5: Conclusion.
Hence, the correct answer is (B) 8.2 g cm$^{-3}$.