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

User Prater
by
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1 Answer

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The density is 3.144 g / cm^3.

Step-by-step explanation:

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

a = 705.2 pm ⇒ In centimeter = 705.2
* 10^-10

Na = 6.023
* 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
* 4) / (6.023
* 10^23)
* (705.2
* 10^-10)

density = 3.144 g / cm^3.

User Henryk Konsek
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4.6k points