dilution factor

Please give us the application note number.

Hey

its 5994-3790

thanks

Step 3 adds the first 10 mL ACN/NH3 - 1 g in 10 mL
Step 5 adds another 10 mL ACN/NH3 - 1 g in 20 mL

Step 10 you take 2 mL out of 20 mL, therefore the DL is 10 = 2/20

if i remove 2ml from the 20ml solution, isnt that the same concentration as the remaining 18ml? assuming proper homogenisation both will have 50mg/ml of gummy

pardon the ignorance, but doesnt dilution usually involve reducing the concentration of an analyte in solution(mg/ml), not just the volume (ml)? for example if i used the 2ml and brought up to 20ml,  that would be x10 dilution? that is generally how i perform dilutions during sample prep. or have i being doing this the wrong way

the paper i look at STEP 3 is just 10ml water. Step 5 ACN/NK3

i believe the quechers salts separates the water (10ml) from the ACN/NH3 (10ml)

Your 2 mL in the original method would only hold 100 mg instead of the 1 g in the beginning.

If you fill up to 20 mL, it is another dilution 1:10, i. e. 1:100 in total; just be sure about the absolute amounts of the sample and what you analyze for your calculations.