Analytical chemistry question?

[irvine_86]

A sample mixture weighing 38.8626g containing only FeCl3 and KCl required 126.22mL of 5.6878M AgNO3 for complete titration of the chloride. How many parts per million (ppm) of potassium is present in the sample?

[Mighty man]

The answer is 2285.6 ppm .... how to get that answer I'm not sure ... but let me know if you find out how to find that answer @MannyRutinel on twitter

Nevermind I found out how to do it:
From the total mass of sample (n is #mol)

38.8626 = (n(FeCl3) x 162.2) + (n(KCL) x 74.5513)

from the titration # mol AgNO3 used gives us the total # mol Cl- = 5.6878 x 0.12622 = 0.717914 mol Cl-

so

0.717914 = 3n(FeCl3) + n(KCl)

use both equations to find n of each constituent. From n(KCL) calculate mass K then ppm is calculated like a mass percentage

ppm K = (mass K/38.8626g) x 10^6 ppm