Question

An element crystallises 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?

विकल्प

• 6.4 × 10²³ atoms

• 12.8 × 10²³ atoms

• 3.2 × 10²³ atoms

• 1.6 × 10²³ atoms

Answer 1

6.4 × 10²³ atoms

Explanation:

Given, Edge length = 500 pm = 500 × 10^-12 m = 5 × 10^-8 cm

Density (d) = 7.5 g cm^-3

Mass(m) = 300 g

For bcc unit, Z = 2

The volume of unit cell (a³)

= (5 × 10^-8 cm)³ = 1.25 × 10^-22 cm³

Volume occupied by each atom

`= (1.25 xx 10^-22 "cm"^-3)/2`

`= 6.25 xx 10^-23 "cm"^-3`

Volume of sample = `"Mass"/"Density"`

`= (300 "g")/(7.5  "g cm"^-3)` = 40 cm³

The number of atoms present in 300 g of the element

`= (40 "cm"^3)/(6.25 xx 10^-23 "cm"^-3) = 6.4 xx 10^23` atoms