Question

An element has a bcc structure with a unit cell edge length of 288 pm. How many unit cells and a number of atoms are present in 200 g of the element? (1.16 × 10²⁴, 2.32 × 10²⁴)

Answer 1

V = a³ = (288 × 10⁻¹⁰ cm)³ = 2.39 × 10⁻²³ cm³

From the relationship:

Density = `(Z xx M)/(N_A xx a^3)`

Where:

Z = 2 (atoms per unit cell for BCC)

M = molar mass in g/mol

N_A = 6.022 × 10²³ mol⁻¹

a³ = volume of unit cell in cm³

Moles = `200/M`

Number of atoms = `200/M xx N_A`

= `200/M xx 6.022 xx 10^23` 

= 2.32 × 10²⁴

M = `(200 xx 6.022 xx 10^23)/(2.32 xx 10^24)`

= 51.9 g/mol

Number of unit cells = `(2.32 xx 10^24)/2`

= 1.16 × 10²⁴