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Flame atomic emission question

  1. Nov 14, 2007 #1
    my question is : How would you make a 1, 3, 5, 7ppm solution from a 100ppm solution?

    do i just just this formula: C1V1 = C2V2 to calculate the amount i need?

    for example, for 5ppm solution => (5ppm) V1 = (100ppm)V2

    so for V1, I just choose my final volume to be whatever i like... like if i was to prepare it in a 100ml volumetric flask, then my V1 would be 100ml.
    so I would need (5ppm*100ml) / 100ppm = 5 ml of the stock 100 ppm soln.

    Is that correct?

    Thank you in advance. =)
     
  2. jcsd
  3. Nov 14, 2007 #2

    Astronuc

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    Staff: Mentor

    If it's ppm then the solution has mgX/kgSol. So 100 mgX/kgSol = 100 ppm.

    If one wants a 1 ppm X solution, take 0.01 (by volume) of 1 kgSol (or 10 gm, which could be 10 ml of aqueous soln) and add 990 gm of solvent (e.g H20), which would give 1 mgX/kgSoln = 1 ppm.

    Or if one uses less volume, e.g. take 0.001 (by volume) of 1 kgSol (1 gm, or 1 ml aq), and add only 99 gm, to get 0.1 mg/0.1 kgSoln.

    See - http://delloyd.50megs.com/photo.html

    http://www.iun.edu/~cpanhd/C101webnotes/aqueoussolns/ppmppb.html

    http://pages.towson.edu/ladon/concas.html

    http://antoine.frostburg.edu/chem/senese/101/measurement/faq/ppm-as-unit-conversion.shtml
     
  4. Nov 14, 2007 #3
    but I am taking from a solution that is already 100 ppm.....
     
  5. Nov 14, 2007 #4

    this sounds right to me
     
  6. Nov 15, 2007 #5

    chemisttree

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    Science Advisor
    Homework Helper
    Gold Member

    It is right, as long as the 5 mL of stock solution is diluted to 100 mL (yielding 5ppm solution). Use 7 mL for the 7 ppm and so forth.
     
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