Question

If the edge length of the unit cell is 705.2 pm, what is the density of KI in g/cm3.

Answer 1

The density  is 3.144 g / cm^3.

Explanation:

If effective number of atom in NaCl type structure, z = 4

a = 705.2 pm ⇒ In centimeter = 705.2 $\times$ 10^-10

Na = 6.023 $\times$ 10^23

density = (molecular weight) (z) / (Na) (a^3)

where molecular weight of KI is 166 g,

           Z represents the atomic number

density = (molecular weight) (z) / (Na) (a^3)              

             = (166 $\times$ 4) / (6.023 $\times$ 10^23) $\times$ (705.2 $\times$ 10^-10)

density = 3.144 g / cm^3.