The substance alpha chromium is found to crystallize in a cubic lattice, with an
Chemistry

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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/cm^{3}, how many Cr atoms are there per unit cell?
Your answer should be an integer:
Hello John I am still working on it. please give me some more minutes
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} cm^{3}
M= D x V
Mcell= 7.561 g/cm3 x 6.624x 10^{22} cm^{3 }
=50.08x 10^12g
Step 2: I am struggling with part two...
let me try to figure out
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} cm^{3}
M= D x V
Mcell= 7.561 g/cm3 x 6.624x 10^{22} cm^{3 }
=50.08x 10^{22}g
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^{23}g)
= 107.3 atoms
= 107 atomsSecure Information
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