25% ammonia means 75% water and 25% ammonia,

If want to 100 ppm( part per million ,i.e.,1 mg/lit) ammonia then add water

1 PPM is just 10,000 times bigger than one percent.

25% ammonia=250000ppm

Xppm*V1=X2ppm*V2

250000ppm X 100=100 ppm X V2

250000 X100=100 XV2

Hence V2=250000 Volume unit, means you add 250000 Volume unit water minus volume you have

For less volume you can take 1/100 part of 25% solution.

1g/ 100mL = 100 ppm solution

ppm= 10^-6 , ppb = 10^-9

Simply do the following:

1. Take 10mL of 25% solution and dilute it to 1000mL mark （1000mL volumetric flask) with water.

2. Take 10mL of the solution prepared in Step 1 and add water to 250m mark (250mL volumetric flask). The concentration of this solution will be 100ppm.

For more accurate information, you should read the product specification. 25% may be the minimum ammonia concentration in the product. The ammonia solution is unstable.

If you want to create standard solutions, you should use certified materials. You can also try to create standard solutions from NH4Cl.

https://www.researchgate.net/post/How_to_make_Ammonium_standard_solution_from_NH4Cl

i) The label concentration of ammonia (aq. sol.) should, in principle, be understood as 25 wt% (as NH3). We can retrieve densities of ammonia (aq. sol.) from (e.g.) Table 2-32, in: R.H. Perry, D.W. Green, J.O. Maloney (Eds.), "Perry's Chemical Engineers’ Handbook", 7th ed., McGraw-Hill, 1997. For 25 wt% (NH3) at 20 ºC; ρ = 0.907 g/cm3 (interpolated). The molecular mass of NH3 is 17.03 g/mol. Hence, the molarity in terms of NH3 would be: 0.25(g NH3 / g aq. sol.)·0.907(g aq. sol. / cm3)·(1000 cm3/dm3)/(17.03 g NH3/mol NH3) = 13.3 M as NH3 (approximate). The label concentration of ammonia should in principle be taken as approximate, namely for quantitative analytical purposes.

ii) It is in principle convenient to dilute this conc. solution by some appropriate dilution factor (f) and then titrate the dilute solution to accurately determine its molarity, which is approximately f·13.3 M. The ammonia dilute sol. can be titrated with standardized strong mineral acid aq. solution. Alternatively, a known excess of standardized strong mineral acid aq. sol. can be added, which excess is then back-titrated with standardized strong base aq. sol., to find the stoichiometric amount of acid previously consumed by the ammonia solution. We can now accurately calculate the the weight (mass) % concentration for the original concentrate solution (~ 25 wt%), by reversing the above exemplified units conversion.

iii) The ppm concentration stands for mg solute / kg solution. Only for sufficiently dilute solutions, the density can be taken as that of water. It is then commonly assumed that the density of water is 1 g/cm3.

Hi Biki,

You may use the web tool to calculate

https://www.alfa-chemistry.com/solution-dilution-calculator.htm

https://www.alfa-chemistry.com/mass-molarity-calculator.htm