An element with molar mass \(2.7 × 10^{-2} kg/mol\) forms a cubic unit cell of edge length \(405 pm\). If the density is \(2.7 × 10^3 kg/m^3\). The nature of the cell is?
Answer & explanation
Correct answer: option 3
The correct answer is option 3. FCC
To determine the nature of the unit cell, we can use the relationship between the density (\( \rho \)), molar mass (\( M \)), and the edge length (\( a \)) for different types of unit cells.
\(\rho = \frac{M}{N_A \cdot a^3}\)
or, \(Z = \frac{\rho N_A a^3}{M}\)
where:
\( \rho \) is the density
\( M \) is the molar mass
\( N_A \) is Avogadro's number
\( a \) is the edge length of the unit cell
Given:
\( M = 2.7 × 10^{-2} \, \text{kg/mol} \)
\( a = 405 \, \text{pm} = 405 \times 10^{-12} \, \text{m} \)
\( \rho = 2.7 × 10^3 \, \text{kg/m}^3 \)
\(N_A = 6.022 × 10^{23}\)
Applying these values in the above equation, we get
\(Z = \frac{2.7 × 10^3 × 6.022 × 10^{23} × (405 \times 10^{-12})^3}{2.7 \times 10^{-2}}\)
or, \(Z = \frac{2.7 × 6.022 405^3 × 10^{-10}}{2.7 ×10^{-2}}\)
or, \(Z \approx 4\)
The lattice with 4 particles is FCC.