An element with molar mass 27 g/mol forms a cubic unit cell with edge length of 405 pm. If the density of the element is 2.7 g/cm^{3}, what is the nature of the cubic unit cell?

#### Solution

**Given:** Edge length (a) = 405 pm = 4.05 × 10^{-8} cm

Molar mass = 27 g mol^{-1}, Density (ρ) = 2.7 g/cm^{3}

**To find:** Nature of cubic unit cell

**Formula:** Density (ρ) = `"M n"/("a"^3 "N"_"A")`

**Calculation:** From the formula,

Density, ρ = `"M n"/("a"^3 "N"_"A")`

∴ `2.7 "g cm"^-3 = (27 "g" "mol"^-1 xx "n")/((4.05 xx 10^-8)^3 "cm"^3 xx 6.022 xx 10^23 "atom mol"^-1)`

∴ n = `(2.7 "g cm"^-3 xx (4.05 xx 10^-8)^3 "cm"^3 xx 6.022 xx 10^23 "atom mol"^-1)/(27 "g mol"^-1)`

= 4.00

∴ Number of atoms in unit cell = 4

Since unit cell contains 4 atoms, it has face-centred cubic (fcc) or ccp structure.

The nature of the given cubic unit cell is face-centred cubic (fcc) or ccp unit cell.