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vesna_86 [32]
3 years ago
6

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?
Chemistry
1 answer:
V125BC [204]3 years ago
4 0

Answer:

Au

Explanation:

For the density of a face-centered cubic:

Density = \dfrac{4 \times M_w}{N_A \times a^3}

where

M_w = molar mass of the compound

N_A= avogadro's constant

a^3 = the volume of a unit cell

Given that:

Density (\rho) = 19.30 g/cm³

a = 0.408 nm

a = 0.408 \times 10^{-9} \times 10^{2} \ cm

a = 4.08 \times 10^ {-8} \ cm

∴

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

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

M_w=197.37 \ g/mol

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

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