Question

The density of KBr is 2.75 g cm⁻³. The length of the edge of the unit cell is 654 pm. Show that KBr has a face-centred cubic structure (N_A = 6.022 × 10²³ mol⁻¹, Atomic masses : K = 39, Br = 80).

Answer 1

Given: ρ = 2.75 g cm⁻³

Edge length of the unit cell (a) = 654 pm = 6.54 × 10⁻⁸ cm

Atomic mass of K = 39 g/mol, Br = 80 g/mol

Molar mass of KBr (M) = 39 + 80 = 119 g/mol

Avogadro’s number (N_A) = 6.022 × 10²³ mol⁻¹

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

⇒ Z = `(rho xx a^3 xx N_A)/M`

∴ a³ = (6.54 × 10⁻⁸)³

= 2.80 × 10⁻²² cm³

∴ Z = `(2.75 xx 2.80 xx 10^-22 xx 6.022 xx 10^23)/119`

∴ Z = `((2.75 xx 2.80 xx 6.022) xx 10^1)/119`

∴ Z = `(46.37 xx 10^1)/119` 

∴ Z = `463.7/119` 

∴ Z = 3.9 ≈ 4

Since Z ≈ 4, this confirms that KBr has a face-centred cubic (FCC) structure.