Question

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?

Answer 1

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).