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

An element has bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm⁻³. For the Bec structure, n = 2

ρ = `"nM"/("a"^3"N"_"A")`

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

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

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

7.2 g = 0.140 M mpl

M = `(7.2  "g")/(0.140  "mol")`

= 51.42 g mol⁻¹

By mole concept, 51.42 g of the element contains 6.023 × 10²³ atom 208 g of the element will contain

= `(6.023 xx 10^23 xx 208)/51.42` atoms

= 24.17 × 10²³ atoms or 2.417 × 10²⁴ atoms