Question

An element occurs in a body-centred cubic structure. Its density is 8.0 g/cm³. If the cell edge is 250 pm, calculate the atomic mass of an atom of this element. (N_A = 6.023 × 10²³)

Answer 1

Given: Z = 2 (BCC)

ρ = 8.0 g/cm³

a = 250 pm = 250 × 10⁻¹⁰ cm = 2.5 × 10⁻⁸ m

N_A = 6.023 × 10²³

To find: Atomic mass (M) = ?

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

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

`M = (8.0 xx 6.023 xx 10^23 xx (2.5 xx 10^-8)^3)/2`

`M = (8.0 xx 6.023 xx 10^23 xx 15.625 xx 10^-24)/2`

`M = (8.0 xx 94.11 xx 10^-1)/2`

`M = (752.88 xx 10^-1)/2`

`M = 75.288/2`

M = 37.644 g/mol