Question

You have a 250ppm solution of K with KCl, you want to prepare a 0.00100 mole of Cl, how many mililiter should be diluted to prepare 1 liter of solution

Answer 1

250 ppm is 250 mg KCl/L or 0.250 g KCl/L or 0.250/molar mass KCl/L = M
Then M = mols x L.
YOu know M and mols needed, solve for L and convert to mL.

Answer 2

thank u

Answer 3

35.4grams/mole*.001mole/2=.0177 grams Cl

If you added to 4ml of the original solutions, you would get
4*250e-6*1g/ml=1e-3 grams of Cl ions, or .001grams
so you want to add not 4 ml, but 4ml(.0177/.001)=70.8ml to make up a volume of 1 liter of solution.