Is my dilution math correct?

[chemist23]

The four standards were first dissolved in water 1:40 (w/v), then mixed 1:1:1:1 (v/v/v/v), filtered (0.45 µm) and finally diluted 1:1 (v/v) with acetonitrile to achieve a 12.5 mg/mL stock solution

Shouldn't the stock solution above be 3.125 mg/ml?

Each of the four standards are different molecules

[James_Ball]

The four standards were first dissolved in water 1:40 (w/v), then mixed 1:1:1:1 (v/v/v/v), filtered (0.45 µm) and finally diluted 1:1 (v/v) with acetonitrile to achieve a 12.5 mg/mL stock solution

Shouldn't the stock solution above be 3.125 mg/ml?

Each of the four standards are different molecules

1:40, then 1:4(one standard to 4 total volume) then 1:1 (one standard to 2 total) so you have 40x4x2 dilution factor which is a 1:320 dilution from the original. If the original weight was 1g then you have 1g\320 = 0.003125g/ml

1:40 = 1g/40ml = 0.025g/ml then dilute to 4ml = 0.00625g/ml then dilute in half with acetonitrile = 0.003125g/ml

Worked through two different ways for the same answer so it seems correct at 3.125mg/ml