Asked by John
An element crystallizes in a body-centered cubic lattice. The edge of the unit cell is 2.77 , and the density of the crystal is 8.65 . Find the atomic weight of the element.
Answers
Answered by
John
55.34g/mol^3
Answered by
Elena
a=2.77 Å= 2.77•10⁻⁸ cm.
The BCC-cell volume is
V=a³=(2.77•10⁻⁸)³ = 2.13•10⁻²³ cm³.
Density ρ =8.65g/cm³ per unit cell.
Avogadro´s number N₀=6.022•10²³ mol⁻¹.
The number of atoms in the BCC unit cell are calculated
as follows:
1 × 1 = 1 center atom
8 × (1/8) = 1 corner atom
=> n=2 total atoms
Atomic weight = N₀•V• ρ/n =
=6.022•10²³•2.13•10⁻²³•8.65/2 = 55.34 g/mol³
The BCC-cell volume is
V=a³=(2.77•10⁻⁸)³ = 2.13•10⁻²³ cm³.
Density ρ =8.65g/cm³ per unit cell.
Avogadro´s number N₀=6.022•10²³ mol⁻¹.
The number of atoms in the BCC unit cell are calculated
as follows:
1 × 1 = 1 center atom
8 × (1/8) = 1 corner atom
=> n=2 total atoms
Atomic weight = N₀•V• ρ/n =
=6.022•10²³•2.13•10⁻²³•8.65/2 = 55.34 g/mol³
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.