Question

An element crystallizes in a bee lattice with cell edge of 500 pm. The density of the element is 7.5 g cm^-3. How many atoms are present in 300 g of metal?

Options

• 3.2 × 10²³ atoms

• 6.4 × 10²³ atoms

• 1.6 × 10²³ atoms

• 12.8 × 10²³ atoms

Answer 1

6.4 × 10²³ atoms

Explanation:

a = 500 pm = 5 × 10^-8 cm

∴ a³ = (5 × 10^-8 cm)³ = 125 × 10^-24 cm³

ρ = 7.5 g cm^-3, m = 300 g, n = 2 (for bcc cell)

`rho = "nM"/("a"^3 "N"_"A")`

`therefore "M" = (rho xx "a"^3 xx "N"_"A")/"n"`

∴ M = `(7.5 xx 125 xx 106-24 xx 6.022 xx 10^23)/2 = 282.3 "g"  "mol"^-1`

282.3 g of metal contains 6.022 × 10²³ atoms

∴ 300 g of metal contains `(6.022 xx 10^23 xx 300)/282.3`

= 6.399 × 10²³ atoms

= 6.4 × 10²³ atoms