At times I get brain-drain when trying to convert results from spectrophotometry in mg/l to mg/kg. could some just give the easiest way involving the volume of digestate and mass of sample.
Usually it will be related with equivalent weight of the liquids. Could you be more specific in describing the case?
So.... if your solvent is water, it is just equal....
If it is not, then, use density of your solvant?
Let's imagin a solvant weight is 2 kg/L then
X mg/L is X/2 mg/kg
By proportionality :
X mg/L is X/d mg/kg with d in kg/L ....
Unless you want to consider that volumes change with solutates, Useless in biology.... so in your case I would say multiply by volume and divide by sample mass.... Maybe a better answer could have been found if I had more detailed description.
@Leduc, let assume a soil mass of of 1g is digested and make up to 25 ml with DD H2O. After spectrophotometry, you get 3.654mg/L of Cu. Kindly convert to mg/kg
Ok, Then... you don't care about the mass of DDH2O (or D2O?) no matter
1g->25mL is 4g->100mL then 40g/L
3.654mg/L is then (3.654)/40 mg/g
that is 3 654 /40 mg/kg that is: 91,35 mg/kg
if you always have 1g of soil in 25 mL then convertion is *1000/40
It could be an exercise of middle school :p
@Thank you Benjamin. I just validated my own method now with your own calculation. I have been using this method- measurement from AAS (mg/L) x Volume of digest/mass of sample. This we get this same answer with your method. Thank you so much.
While working with soils, we generally do extraction of elements with Mehlich-3 extracting solution. In your case, Let us say (or assume) — that you are digesting 3 grams of soil sample, and making it up to 30 mL with the Mehlich solution.
- If you have 3g soil in 30mL of solution, then you have 100g soil per liter of solution
On the other hand, after digestion and analysis of solution — you found the Phosphorous concentration to be 1 mg/l.
- You have 1mg P per liter of solution
On comparing the above two bullet points, you can conclude that, you have 1mg P in 100g soil (or) 10mg in 1kg (or) 10 mg/kg
If it was a sol, for exemple you did leaching or extraction, you have to convert the unity from mg/l to mg/Kg, so usually we use this method:
for example (leaching):
100g ->1L
the concentration of nitrite is : 2mg/l
so you do 2/100 (mg/g), after that you get 0.02*1000 to have 20mg(nitrite)/kg sol
If it was a sol, for exemple you did leaching or extraction, you have to convert the unity from mg/l to mg/Kg, so usually we use this method:
for example (leaching):
100g ->1L
the concentration of nitrite is : 2mg/l
so you do 2/100 (mg/g), after that you get 0.02*1000 to have 20mg(nitrite)/kg sol
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