Question

Determine the density of Cesium chloride which crys tallizes in BCC type structure with the edge length 412.1 pm. The atomic masses of Cs and Cl are 133 and 35.5 respectively.

Answer 1

Given: Edge length (a) = 412.1 pm = 4.12 x 10^-8 cm

Molar mass = 133 + 35.5 = 168.5 g mol^-1

To find: Density (d)

Formulae: Mass of one molec = Molar mass/Avogadro number

Volume of unit cell = a³

Density = Massof unit cell/Volumeof unit cell

Calculation: In the bcc type unit cell of CsCl, there is one Cs⁺ ion at the body centre
position and 8 Cl^- ions are at the 8 corners

∴ Number of Cs⁺ in unit cell = 1

Number of Cl^- in unit cell = 1/8 x 8 = 1

Hence, the unit cell contains one CsCl molecule

Mass of one CsCl molecule = Molar mass/Avogadro number = 168.5 g mol^-1/6.023 x 10²³ mol^-1

= 2.798 x 10^-22 g

∴ Mass of unit cell = 1 x 2.798 x 10^-22 g = 2.798 x 10^-22 g

Volume of unit cell = a³ = (4.12 x 10^–8 cm)³ = 6.993 x 10^-23 cm³

∴ Density = Massof unit cell/Volumeof unit cell = 2.798 10^-22/6.993 x 10^-23 cm³ = 4.0 g cm^–3