Question

An element ‘X’ having atomic mass 60 has density 6.23 g cm⁻³. The edge length of its unit cubic cell is 400 pm. (N_A = 6.02 × 10²³ mol⁻¹).

What is the type of unit cell known as?

Options

• Body centred cubic

• Face centred cubic

• Simple cubic

• Side centred cubic

Answer 1

Face centred cubic

Explanation:

Given: Atomic mass (M) = 60 g/mol

Density (ρ) = 6.23 g/cm³

Edge length (a) = 400 pm = 4.00 × 10⁻⁸ cm

Avogadro’s number (N_A) = 6.02 × 10²³ mol⁻¹

We have to determine the type of unit cell by calculating the number of atoms per unit cell (Z).

`rho  = (Z xx M)/(a^3 xx N_A)`

`Z = (rho xx a^3 xx N_A)/M`

`Z = (6.23 xx (4 xx 10^-8)^3 xx 6.02 xx 10^23)/60`

`Z = (6.23 xx 64 xx 10^-24 xx 6.02 xx 10^23)/60`

`Z = (6.23 xx 385.28 xx 10^-1)/60`

`Z = (2400.2 xx 10^-1)/60`

`Z = 240.02/60`

Z = 4

∴ This is a Face-Centred Cubic (FCC) unit cell.