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Finding number of atoms per cm^3 of zinc?

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Zinc has a density of 7.17 Mg/m^3. Calculate (a) the number of Zn atoms per cm^3, (b) the mass of a single Zn atom and (c) the atomic volume of Zn.


    2. Relevant equations
    atomic mass of zinc = 65.39 g/mol


    3. The attempt at a solution

    For part (a) I use the fact that zinc has atomic mass of 65.39 g/mol. So dividing 7,170,000 (g/m^3) by 65.39 (g/mol), you get 109,649.79 (mol/m^3). Then we divide that by 10^6 to get .10964979 mol/cm^3. Then we multiply that by avagadro's number to get 6.603*10^22 atoms/cm^3. Is this right?

    For part (b) we know that one cm^3 of zinc has 7,170,000/10^6 grams of mass. AKA 7.17 grams of mass. And we also know from part (a) that one cm^3 of zinc has 6.603*10^22 atoms in it. So each atom has 7.17/6.603*10^22= 1.0859*10^-22 grams of mass. Is that right?

    For part (c) we use the fact from part (a) once cm^3 of zinc has has 6.603*10^22 atoms in it. So that mean an individual zinc atom has a volume of 6.603*10^-22 cm^3. Is this right?
     
  2. jcsd
  3. Sep 12, 2013 #2
    Looks right.
     
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