Determine the density of CsCl which crystallizes in a bcc types structure with an edge length 412.1 pm.

Explanation

Structure of cesium chloride is body centre cubic type in which `Cs^(+)` placed within centre and `Cl^(-)` placed at cover of lattice.
Number of `Cl^(-)` ions per unit cell `= 8 xx (1)/(8) = 1 `
Number of `Cs^(+)` ions per unit cell =1
So there is only one molecule of CsCl in a lattice
Mass of one molecule of CsCl
`= ( "Molar mass of CsCl")/("Avogadro's number")`
`= ( 168.5)/(6.02 3 xx 10^(23))`
`= 2.8 xx 10^(-22) g `
Volume of lattice `= a^(3)`
`= ( 412.1 xx 10^(-12))^(3)` or
`= 7 xx 10^(-23) cm^(3)`
Density of CsCl `= ( 2.8 xx 10^(-22))/( 7 xx 10^(-23))`
`= 4g //cm^(3)`