An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density of the element is 19.3 g/cm3 , what is the identity of the element?

Respuesta :

Answer:

Au

Explanation:

For the density of a face-centered cubic:

[tex]Density = \dfrac{4 \times M_w}{N_A \times a^3}[/tex]

where

[tex]M_w[/tex] = molar mass of the compound

[tex]N_A=[/tex] avogadro's constant

[tex]a^3 =[/tex] the volume of a unit cell

Given that:

Density [tex](\rho)[/tex] = 19.30 g/cm³

a = 0.408 nm

a = [tex]0.408 \times 10^{-9} \times 10^{2} \ cm[/tex]

a = [tex]4.08 \times 10^ {-8} \ cm[/tex]

[tex]19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}[/tex]

[tex]M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}[/tex]

[tex]M_w=197.37 \ g/mol[/tex]

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

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