I think I got lost somewhere between the milli-moles & the 500 PPM.
So is the stoichiometric amount of Maltose & Dextrose I mentioned correct for 20 PPM?
Dextrose = 40 mg per Liter (0.04 gm) ?
Maltose = 80 mg per Liter (0.08 gm) ?
Silver atom weight / molar weight = 107.87
Glucose / Dextrose atom weight / molar weight = 180.16
Maltose atom weight / molar weight = 342.30
So...
1 liter 20 ppm reduced with Glucose/Dextrose:20 mg silver / atom weight of silver 107.87 = 0.1854 milli-mol
0.1854 milli-mol silver * 180.16 atom weight of glucose = 33.40 mg glucose/dextrose
And since 1 molecule of glucose will reduce 1 molecule of silver oxide, which contains 2 silver atoms, only half that weight of glucose/dextrose is needed...
33.40 / 2 = 16.7 mg, which is 0.0167 gram, we round up to 0.02 gram (or more).
1 liter 20 ppm reduced with Maltose:20 mg silver / atom weight of silver 107.87 = 0.1854 milli-mol
0.1854 milli-mol silver * 342.30 atom weight of maltose = 63.46 mg maltose
And half of that...
63.46 / 2 = 31.73 mg = 0.03173 gram, which is rounded up to 0.04 gram (or more).