1 ppm = I mg/liter

however, it is only 50% AI so 1 ppm AI = 2 mg/liter

A 500 ppm solution at 50% AI would be a 1 g/l (=1000 mg/liter).

There are a bunch of problems to consider:

What is the field rate, or more precisely how does the field rate translate into your laboratory test. In the field the pesticide in aqueous solution will be applied as an atomized spray. This will create discrete deposits on the foliage. The toxicant concentration, deposit size, and number of deposits will interact to influence efficacy. Most people don't want to deal with this complexity, but it has significant impacts on lab-to-field performance.

that's right timothy 1g of the compound (whatever) in 1 lit of water = 500PPM

Hi, first corectiion factor; 100/50=2

then for 100 cc:

2*100*500=100000 microgram per 100cc solution. it would be 0.1 gr per 100 cc. now you have stock of 500 ppm in volume of 100 cc. 

Dissolve 1g of compound in 1L of water. Now you have 1L of 500 ppm pesticide solution.

Sarah,

I realize ppm is still used but it is not an SI unit. 

The scientific journals I deal with will make you change ppm to g/L or something similar.

I suggest just using SI if you want to publish.

I would suggest molarity. It is standard in any chemistry class. However, it is not used much in pesticide literature.

500 ppm translates to 500 mg/L. Then you weigh 500mg of the solid pesticide, dissolve it in a small volume of distilled water and make the solution up to the one litre mark on a measuring cylinder. I hope that the pesticide is soluble in water, otherwise you will need to use small volume of tween 40/80 to make initial water solubility. SORRY I JUST SAW THE 50% BIT, THEN YOU WILL NEED 1000MG/L.

@David

That works great, except that we need a 500 ppm active ingredient from a material that is 50% active ingredient. So in your example 500 mg of solid gives only 250 mg active and results in a 250 ppm solution.

A surfactant like tween may help. It also can help to dissolve the active ingredient in something like acetone first (being careful that the solvent does not react with the toxicant). One then adds a small quantity of acetone (or tween, or whatever is used) to the water used to treat the control. This allows you to factor out the toxicity (if any) of the material used to aid is dissolving the pesticide.

For that matter, you should consider that the "inert" ingredients in pesticide formulations are not biologically inactive, and any adjuvants used also have biological activity. Typically you are testing the active ingredient + formulation components and not just the active ingredient. Mostly this is good because a grower will not have the option of separating the two. Scientifically, it is important to keep this distinction in mind.

Thanks Tim. I did not see the 50% AI bit earlier. Your positions are excellent. Many thanks.

Dr Timothy has suggested the correct procedure. Be sure about the active ingredient.

Dr. Prasad, The active ingredient is certainly very important because of other inclusions in pesticide formulations that are not active towards actual performance and these need be eliminated in final concentration. Thanks for your contribution

@David

The "non-active" ingredients are of many classes. Some may be filler, but most are there to improve product performance. This includes alter water quality, improved penetration into target organism, improved deposition, or reduced degradation rates. All of these will change the potency of the active ingredient. It is a very different experiment when you eliminate the formulation components from the final consideration. Mostly, one tests the formulation, not just the active molecule.

If we would like to prepare 0.0001 ppm from 57% then what will be the procedure

this concentration is very little and have not any effect

LD50 for malathion 2500 mg/kg

ADI= 0.01 mg/kg/day

0.0001 ppm = 0.0001/ 1000ml

57 in 1000

0.0001 in X

X= 1000x0.0001/57

0.1/57/ 1000ml.

for saturation solubility study, i have taken 100mg drug in excipient, after mixing i have filtered out and it was 1.2ml. the from that 1.2ml, i have taken 1ml and diluted it to 1ml of ethanol. from this 2ml of mixture i have taken 0.5ml and diluted it to 10ml. i have taken the absorbance by UV. actually i cant understand what is the ppm of stock solution ?

You can weight 500 mg or 0.500 g of solid pesticide and dissolve it in one liter of solvent such as distilled water.