The substance alpha chromium is found to crystallize in a cubic lattice, with an edge length of 871.7 pm. If the density of solid alpha chromium is 7.561 g/cm3, how many Cr atoms are there per unit cell?
Hello Jay I am almost through but this one is a little bit difficult
this is how you go about the question.
Step1: since the density is given in
g/cm3, use the edge length to calculate the volume of the cell in cm3, then use
the density to calculate in one unit cell.
For a cubic cell the volume is the
edge length cubed. 1pm=10^-12m =10^-10cm
Vcell= (871.7x10-10 cm)3
= 6.624x 10-22 cm3
M= D x V
Mcell= 7.561 g/cm3 x 6.624x 10-22 cm3
Step 2: I am struggling with part two
Step1: since the density is given in g/cm3, use the edge length to calculate
the volume of the cell in cm3, then use the density to calculate in one unit
Step 2: use the molar mass of the element to calculate the mass of 1 atom. Then
divide the mass of the unit cell by the mass of one atom to get the number of
atoms in the cell.
We take the mass to be
M atom = (28.09g/mol) x (1
mol/ 6.022x 10-23 atoms = 4.665x10-23 g/atom
Number of atoms = (50.08x 10-22g/
cell) x (1 atom/ 4.665 x 10-23g)
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