Answer:
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
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:
[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]
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
[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]
[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
