This question is first year chemistry. Check any analytical chemistry textbook and you will find it.

Dear C.A. (Kees) Kan thanks for your kind reply and am agree with your reply but i did not study chemistry in my graduation that's by facing these kind of problems.

please recommend a book

My books are too old to recommend. Search the net or consult your chemistry department.

The general formula for making dilutions out of the stock solution is C1.V1=C2.V2

C= concentration, V= volume

In your case, if you mind to make 100 ml of the Solution#2 (100 ppm) out of your stock which is 1000 ppm, you need to solve the formula for V1. The formula after putting the numbers will look like this:

1000*V1= 100*100

V1= 10ml

So you need to get 10 ml of your stock solution (1000ppm) and add 90 ml distilled water(or any other solvents) to that. Now you have solution#2 which contains 100 ppm of your material.

Dilute your solution#2 with 1:1 ratio (50 ml from solution 2+ 50 ml of solvent= 100 ml of solution#3) with a proper solvent to give you 50 ppm concentration or solution#3.

To make solution#4 which contains 10 ppm of your desire stuff, dilute your solution#2 with 1:10 ratio( 10 ml solution#2 + 90 ml of the solvent)

the final concentration could be made by diluting your solution#3 with a ratio of 1:100 (1 ml of solution#3 + 99 ml of the solvent).

In this way, you will have below concentration:

Stock solution or solution#1= 1000 ppm

Solution#2= 100 ppm

Solution#3= 50 ppm

Solution#4= 10 ppm

And Solution#5= 0.5 ppm

Stock Concentration = 1000ppm; 1gm of metal salt in 1000ml of solvent system

{Deionized water (dw)}

Formula C1V1=C2V2; where C1 = stock concentration,

V2= volume to be taken for stock for working solution preparation,

C2= required concentration

V2 = required volume

using this formula you can prepare the working solutions.

plz have a look of attached file for calculation

all the best

Bhavya G. I disagree that 1 gr of metal salt in one l of water will yield 1000 ppm of the metal in the solution.

You have to take into account the counter ion of e.g. Pb and depending on the salt, the amount of crystal water in the salt. So the calculation depends on the salt used.

thank you@ Amir Ahmadpour for sharing the information.

Thank you@ C.A (Kees) Kan for providing detailed information.

Je vais m'exprimer en français

Si c'est une solution stock de 1000ppm de cadmium et de plomb a partir d'un produit commercial de cadmium et un produit commercial de plomb

Alors il vous faudra calculer la masse de la solution commerciale de cadmium à prélever pour obtenir 1000 ppm de cadmium et il faudra en faire de même pour le plomb

Soit m = CMV

C (1000 ppm) M (masse moléculaire du cadmium) et V (le volume à préparer soit 1 L si on veut)

Idem pour le plomb

C(1000 ppm) M (masse moléculaire du; plomb) et V (le volume à préparer Soit 1L si on veut)

On obtient une masse m1 et une masse m2 qu'on met dans une fiole de 1L qu'on complète jusqu'au trait de jauge.

Ainsi, on obtient une solution mère de 1000 ppm de cadmium et de plomb.

A partir de cette solution, on fait des dilutions

Soit C initial x V initial = C final x V final

D'où V initial = (C final x V final ) / C initial

avec C initial (1000 ppm) , C final (100 ppm ou 50 ppm) et V final (le volume final à obtenir)

Voilà ce que je peux apporter comme contribution et je reste à l'écoute

Merci

For preparing 1000 ppm of Lead stock solution , take 1.6 g of lead nitrate and dissolved in 1 lit of deionised water. For other concentrations of 100, 50, 10 and 0.5 ppm , the dilution technique is good option.

Just like C A (Kees) Kan, mentioned it depends on the nature of the salt. But generally, you need to find the gram equivalent of the salt. Example, gram equivalent of Pb = M wt of Pb(NO3)2 * 100 all over Atomic wt of Pb * %purity.

I have prepared the ppm solution by dissolving 1g of solute in 1000ml of solvent. There is one condition which is applicable as the molar mass of solute is varying for different solutes so one has to follow some equation prior the process of dissolving.