The question is incomplete, here is the complete question:
Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 456.2 pm, what is the density of CsI in
![g/cm^3](https://img.qammunity.org/2021/formulas/physics/middle-school/10snjt0scu2x7abt0uifdskb703d9bptvv.png)
The image is attached below.
Answer: The density of CsI is
![9.09g/cm^3](https://img.qammunity.org/2021/formulas/chemistry/college/g81mhnilrhnrd8r8vlokr6fsabemcctwzn.png)
Step-by-step explanation:
To calculate the density of metal, we use the equation:
![\rho=(Z* M)/(N_(A)* a^(3))](https://img.qammunity.org/2021/formulas/chemistry/college/yjrfpwvf6kb4dasljsxig3d849folkihg9.png)
where,
= density
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of CsI = 259.8 g/mol
= Avogadro's number =
![6.022* 10^(23)](https://img.qammunity.org/2021/formulas/chemistry/middle-school/g7xgyg45ozrq4pcvaillv1vuj3s44s64vb.png)
a = edge length of unit cell =
(Conversion factor:
)
Putting values in above equation, we get:
![\rho=(1* 259.8)/(6.022* 10^(23)* (456.2* 10^(-10))^3)\\\\\rho=9.09g/cm^3](https://img.qammunity.org/2021/formulas/chemistry/college/tfjxs6uj5i42yjky6dcg891pyino1bp8lc.png)
Hence, the density of CsI is
![9.09g/cm^3](https://img.qammunity.org/2021/formulas/chemistry/college/g81mhnilrhnrd8r8vlokr6fsabemcctwzn.png)