Respuesta :
Answer:
Face Centered Cubic
Explanation:
In this question our strategy should be to determine the number of atoms in the unit cell since we know that the cubic structure can be:
primitive = 1 atom/ unit cell )
body centered = 2 atoms/ unit cell )
face centered = 4 atoms/ unit cell )
We know the density ( d= m/V ) , the volume can be computed and the molecular weight, so we have all the information required.
m unit cell = # atoms in the unit cell x mass of 1 atom
Need to convert the lattice parameter, a, in Angstroms to cm for unit consitency with the density.
1 A = 10⁻⁸ cm
a = 6.0849 A x 1x 10⁻⁸ cm/A
a = 6.0849 A x 1.x 10⁻⁸ cm
⇒ V = a³ = ( 6.0849 x 10⁻⁸ cm)³ = 2.3 x 10⁻²² cm³
m of the unit cell = d x V = 2.6 g/cm³ x 2.3 x 10⁻²² cm³ = 5.98 x 10⁻²² g
We need to convert the atomic weight per mol to atomic weight per atom:
87.62 g/ mol x (1 mol / 6.022 x ²³ atom ) = 1.45 x 10⁻²² g/atom
Then this mass of 5.98 x 10⁻²² g must be equal to
5.98 x 10⁻²² g = # atoms/ unit cell x 1.45 x 10⁻²² g/atom
4.11 = # atoms/ unit cell
Rounding off to integer # atoms/unit cell = 4
Therefore the crystal structure of the cell is face centered cubic.
The crystal structure of the metal should be considered as the Face Centered Cubic.
Identification of the crystal structure:
Here first determine the no of atoms in unit cell
So,
primitive = 1 atom/ unit cell )
body centered = 2 atoms/ unit cell )
face centered = 4 atoms/ unit cell )
Now
1 A = 10⁻⁸ cm
So,
a = 6.0849 A x 1x 10⁻⁸ cm/A
a = 6.0849 A x 1.x 10⁻⁸ cm
Now
V = a³ = ( 6.0849 x 10⁻⁸ cm)³ = 2.3 x 10⁻²² cm³
So,
m of the unit cell
= d x V = 2.6 g/cm³ x 2.3 x 10⁻²² cm³
= 5.98 x 10⁻²² g
Now
Here we are required to convert the atomic weight per mol to atomic weight per atom
So,
87.62 g/ mol x (1 mol / 6.022 x ²³ atom ) = 1.45 x 10⁻²² g/atom
So, the mass of 5.98 x 10⁻²² g should be
5.98 x 10⁻²² g = # atoms/ unit cell x 1.45 x 10⁻²² g/atom
4.11 = # atoms/ unit cell
= 4
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