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. (N₀ = 6.023 × 1023)

Answer 1

Z = 2(BCC)

ρ = 8.0 g/cm³

a = 250 × 10^-10 cm

N₀ =6.022 × 10²³

A = ?

We know that ρ = `("Z" xx "A")/("N"_0 xx "a"^3)`

A = `(rho xx "N"_0 xx "a"^3)/"Z"`

`= (8.0 xx 6.022 xx 10^23 xx (250 xx 10^-10)^3)/2`

A = 94 amu