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 [tex]g/cm^3[/tex]
The image is attached below.
Answer: The density of CsI is [tex]9.09g/cm^3[/tex]
Explanation:
To calculate the density of metal, we use the equation:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of CsI = 259.8 g/mol
[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
a = edge length of unit cell = [tex]456.2pm=456.2\times 10^{-10}cm[/tex] (Conversion factor: [tex]1cm=10^{10}pm[/tex] )
Putting values in above equation, we get:
[tex]\rho=\frac{1\times 259.8}{6.022\times 10^{23}\times (456.2\times 10^{-10})^3}\\\\\rho=9.09g/cm^3[/tex]
Hence, the density of CsI is [tex]9.09g/cm^3[/tex]
