(b) Using the Crystallographer's formula to determine the density (in g/cm³) of Po: ρ = Σ Ζ * Μ Vell* N You must know that a primitive cubic cell has Z = 1 Since Po is a metal, the value of "i" = 1, since it's the only entity! MM of Po= 208.998 g/mol Vcell (in cm³) = a³ Vcell (3.36 x 10-8 cm)³ = 3.793 x 10-23 cm³ N = 6.022 x 1023 atoms/mol Thus: p = (1 atom/cell) (208.998 g/mol) = 9.16 g/cm³ (3.793 x 10-23 cm³) (6.022 x 1023)

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Chapter11: States Of Matter; Liquids And Solids
Section: Chapter Questions
Problem 11.18QP: What is the coordination number of Cs in CsCl? of Na in NaCl? of Zn2 in ZnS?
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(b) Using the Crystallographer's
formula to determine the density (in g/cm³) of Po:
ρ = Σ Ζ * Μ
Vell* N
You must know that a primitive cubic cell has Z = 1
Since Po is a metal, the value of "i" = 1, since it's the only entity!
MM of Po= 208.998 g/mol
Vcell (in cm³) = a³
Vcell (3.36 x 10-8 cm)³ = 3.793 x 10-23 cm³
N = 6.022 x 1023 atoms/mol
Thus: p = (1 atom/cell) (208.998 g/mol) = 9.16 g/cm³
(3.793 x 10-23 cm³) (6.022 x 1023)
Transcribed Image Text:(b) Using the Crystallographer's formula to determine the density (in g/cm³) of Po: ρ = Σ Ζ * Μ Vell* N You must know that a primitive cubic cell has Z = 1 Since Po is a metal, the value of "i" = 1, since it's the only entity! MM of Po= 208.998 g/mol Vcell (in cm³) = a³ Vcell (3.36 x 10-8 cm)³ = 3.793 x 10-23 cm³ N = 6.022 x 1023 atoms/mol Thus: p = (1 atom/cell) (208.998 g/mol) = 9.16 g/cm³ (3.793 x 10-23 cm³) (6.022 x 1023)
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