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Salt crystal forms after 3 days of evaporation

  1. Nov 1, 2008 #1
    If a salt crystal forms after 3 days of evaporation, and the crystal mass is 30 mg, how many Na+ and Cl- ions was added to the crystal each second (average). How do I solve this?
     
  2. jcsd
  3. Nov 1, 2008 #2

    symbolipoint

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    Re: Saltcrystal

    This is essentially a units conversion problem. You want to change from milligrams per days to atoms per second. In this case, whether "ions" or "atoms" does not change the units conversion needed. You will need to first to determine, calculate the formula weight of sodium chloride.
     
  4. Nov 1, 2008 #3
    Re: Saltcrystal

    Ok, so I have [tex]MM_{NaCl}=58.44277 \text{g/mol}[/tex], and I guess I have will have to find the amount of moles; [tex]n=\frac{30\times 10^{-3}}{58.44277}=5,1332\times 10^{-4}[/tex]. So far so good? Would the next thing be to do [tex]\frac{5,1332\times 10^{-4}}{3*24*60^2}[/tex], to get the number of moles crystallized per second?
     
  5. Nov 1, 2008 #4

    symbolipoint

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    Re: Saltcrystal

    Your work seems good; now remember since you wanted "ions", the actual count of ions in the amount of time, you need to use 1 mole = 6.022*10^23 atoms or whatever individual countable item you are counting.
     
  6. Nov 2, 2008 #5
    Re: Saltcrystal

    Excellent! So, I got [tex]1,9804\times 10^{-9}\mbox{ moles/second}[/tex]. So the amount of Na and Cl that crystallizes each second would be [tex]6,0221415\times 10^{23}\times 1,9804\times 10^{-9}=1,1926\times 10^{15}[/tex]. I guess that since the mole ratio is 1:1, [tex]Na^{+}=\frac{1,1926\times 10^{15}}{2}\mbox{, } Cl^{-}= \frac{1,1926\times 10^{15}}{2}[/tex], right?
     
  7. Nov 2, 2008 #6

    symbolipoint

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    Re: Saltcrystal

    Instead of confirming or denying your results in post #5 directly, try reworking everything on paper (for convenience) and include ALL units in your arrangement; this will help you be clear about your arrangement and you can perform your units analysis.
     
  8. Nov 2, 2008 #7
    Re: Saltcrystal

    Ok, I did, and I ended up with getting atoms/second, which I guess is..right?
     
  9. Nov 3, 2008 #8
    Re: Saltcrystal

    i think you don´t have to divide the sum of atoms by 2 to get the number of Na+ or Cl- atoms, because if you have, lets say 1mole of NaCl, then it has 1mole of Na+ and 1mole Cl- atoms, which means twice as many particles as in NaCl.
     
  10. Nov 3, 2008 #9

    symbolipoint

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    Re: Saltcrystal

    vaazu, you are correct; one could first view the count of units of NaCl, and then multiply by two, since there are 2 atoms in one formula's atomic combination for NaCl.
     
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