Question

Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 428.7 pm, what is the density of CsBr in g/cm3.

Answer 1

`\mathbf {density \ d =4.4845 \ g/cm^3}`

Step-by-step explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

`\mathtt{ density \ d = (z * molar\ mass \ (M))/( edge \ length \ (a) \ * avogadro's \ number \ (N))}`

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

`\mathtt{ density \ d = (1 * 212.81)/((4.287 * 10^(-8))^3 * 6.023 * 10^(23))}`

`\mathtt{ density \ d = ( 212.81)/(47.4540533)}`

`\mathbf {density \ d =4.4845 \ g/cm^3}`