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?

FCC

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.