# Homework Help: Flame atomic emission question

1. Nov 14, 2007

### higherme

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?

2. Nov 14, 2007

### Astronuc

Staff Emeritus
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 [Broken]

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

http://antoine.frostburg.edu/chem/senese/101/measurement/faq/ppm-as-unit-conversion.shtml

Last edited by a moderator: May 3, 2017
3. Nov 14, 2007

### higherme

but I am taking from a solution that is already 100 ppm.....

4. Nov 14, 2007

### eli64

this sounds right to me

5. Nov 15, 2007

### chemisttree

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.