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.
2 answers
55.34g/mol^3
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
1 answer