Question

An element has bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm⁻³. How many atoms are present in 208 g of the element?

Answer 1

The bcc structure, n = 2

`ρ = (n xx M)/(a^3 xx N_A)`

`7.2  g  cm^-3 = (2 xx M)/((288 xx 10^-10  cm)^3 xx (6.023 xx 10^23  mol^-1))`

`7.2  g  cm^-3 = (2 xx M)/((23887872 xx 10^-30  cm^3) xx (6.023 xx 10^23  mol^-1))`

`7.2  g  cm^-3 = (2 xx M)/((2.38 xx 10^-23  cm^3) xx (6.023 xx 10^23  mol^-1))`

`7.2  g  cm^-3 = (2 xx M)/(14.33  cm^3  mol^-1)`

`((7.2  g  cm^-3) xx (14.33  cm^3  mol^-1))/2 = M`

M = `(103.176  g  mol^-1)/2`   

M = 51.58 g mol⁻¹

By mole concept, 51.58 g of the element contains 6.023 × 10²³ atom 208 g of the element will contain

= `(6.023 xx 10^23 xx 208)/51.58` atoms

= `(1252.784 xx 10^23 )/51.58` atoms

= 24.2 × 10²³ atoms ≈ 2.42 × 10²⁴ atoms