[Baileys]

I need to find the original conc of As in a sediment sample, after carrying out these dilutions:

Of the sediment, 0.5g was taken, treated with various chemicals, and then made up to the mark in a 50mL flask.

A 4mL aliquot of this was taken and transfered into another 50mL flask.

A 5mL aliquot of this was transfered into a 100mL flask, which was made up to the mark.

The absorption reading obtained for As was 2.5mg L-1.


This is what I have done so far, where I have worked backwards in the dilutions.
Please let me know if I am on the right lines or not.

Total As in 100mL flask = 2.5 x 100/1000 =  0.25mg
Total As in the 5mL aliquot = 0.25mg
Total As in 50mL aliquot = 0.25 x 50/5 = 2.5mg
Total As in 4mL aliquot = 2.5mg
Total As in 50mL flask = 2.5 x 50/4 = 31.25mg

Conc in sediment = 31.25/0.5 = 62.5mg g-1

I won't bother converting it into mg/kg. I just want to make sure I have the right jist here

[JGK]

AS level 2.5 mg/L

total dilution factor 4 mL in 50 (x 12.5) followed by 5 mL in 100 (x 20) = x 250

therefore the original 50 mL sample has a conc of 625 mg/L and 50 mL will contain 31.25 mg

31.25 mg is present in 0.5 g of the sediment = 62.5 mg/g

[Golden_4_Life]

As (no pun intended) J G  K your calculations are both clear and lucid. Is this done on a relative % basis or ''as is".

[JGK]

[AS] is basis - definitely