Canyou show me how to solve?

Question

an element crystalizes in a body center cubic lattice. The edge of the unit cell is 2.86angstrims, and the density of the crystal is 7.92g/cm3. What is the atomic weight of the element?

Answer 1

A body-centered cubic lattice has an atom at each of 8 corners of the cell and an atom in the center of the cell. The fraction of an atom that occupies a corner of a unit cell is 1/8 and the fraction of the center atom in the unit cell is, of course, 1. So,

8 corner atoms * (1/8) + 1 center atom = 2 atoms in the unit cell

Then, density = mass / volume
density = 7.92 g/cm^3
mass = x amu (what you are looking for)
volume = (2.86 Angstroms)^3 (length times width times height of unit cell)

amu = density/volume = (7.92 g/cm^3)/(2.86 A)^3

Then to convert amu to grams, divide the number of amu from above step by Avogadro's number, 6.02 x 10^23 amu. This gives the answer in grams.
That oughta do it!

Answer 2

In a body-centered cubic latice, there is one atom at the center of the cube and 1/8 atom at each corner, shared with all 8 of the adjacent cubes. There are 8 corners per cube, so 2 atoms per cell (8 x 1/8 = 1 and + 1 =2).

1 cell/2atoms x 6.02x10^23atoms/1mol x (2.86x10^-8cm)^3cu cm/cell x 7.92g/1cu cm =

1A = 10^-8cm

6.02 x 10^23 is Avogadro's njmber.