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1 mol of Al atoms = 6.022e23 atoms
Volume of unit cell = (4.05 angstrom)^3 = 66.43 angstrom^3
# of atoms per unit cell (fcc) = 1/8*8 + 1/2*6 = 4 atoms
# of unit cells = 6.022e23/4 = 1.5055e23 unit cells
total volume = (1.5055e23)*(66.43 angstrom^3) = 10 cm^3where did the 111.0412 au^3 (whatever au^3 is) and 6.7 cm^3 come from?
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