dilution factor

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

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.

yes,i get that the absolute value would reduce. Im just confused as this is not how i have done other samples. in chemstation the sample amount is usually given as a concentration (ug/ml) rather than an absolute mass (ug)

just to be clear, "1:10 dilution" ( 2ml to 20ml) is different from "dilution factor x10" (2ml out of 20ml), is this what you are saying?

sorry if this seems obvious, im just tring to understand

yes,i get that the absolute value would reduce. Im just confused as this is not how i have done other samples. in chemstation the sample amount is usually given as a concentration (ug/ml) rather than an absolute mass (ug)

just to be clear, "1:10 dilution" ( 2ml to 20ml) is different from "dilution factor x10" (2ml out of 20ml), is this what you are saying?

sorry if this seems obvious, im just tring to understand