Question

A face-centred cubic element (atomic mass = 60) has a cell edge of 400 pm. What is its density?

पर्याय

• 0.623 g cm⁻³

• 6.23 g cm⁻³

• 62.3 g cm⁻³

• 0.0623 g cm⁻³

Answer 1

6.23 g cm⁻³

Explanation:

Given: Structure: Face-Centred Cubic (FCC) ⇒ Z = 4 atoms/unit cell

Atomic mass (M) = 60 g/mol

Edge length (a) = 400 pm = 4 × 10⁻⁸ cm

Avogadro’s number (N₀) = 6.022 × 10²³ mol⁻¹

`rho = (Z * M)/(a^3 * N_0)`

`rho = (4 xx 60)/(4 xx 10^-8)^3 xx 6.022 xx 10^23`

`rho = 240/(64 xx 10^-24 xx 6.022 xx 10^23)`

`rho = 240/(384 xx 10^-1)`

`rho = 240/38.4`

ρ = 6.25 g cm⁻³ ≈ 6.23 g cm⁻³