I really wouldn't be too surprised if there's a way of getting 90%+ purity relatively easily, but I'm no chemist. I guess I'll have to keep searching.
I am actually a chemsit lol. I'm mastering in Analytical Chemistry so I only have a basic knowledge of undergrad inorganic synthesis, to my knowledge in the sysnthesis experiments I done sometimes getting such a high purity from like initial synthisis steps is just physically not possible, like 30% purity might be a "good" yield for a given method.
There might be like purification methods such as recrystalisation actually, that could let you increase purity but youd maybe have to see what yield you get. Often what seems like tonnes of unpurified product can boil down to small amounts when purified to a hihh degree. You could maybe research into recrystalisation methods though, in theory if you can get a good solvent it could probably just be done through like coffee filters if you worked fast (we sometimes use fancy vacuum filtration techniques in the lab), you can google the method and that'll make sense. Again I very well could be talking out my ass though I always hated synthesis experiments.
Also one more thing I remembered after commenting the first time, I think I remember reading SN is pretty air sensitive, like it turns back to sodium nitrate if exposed to air for long enough, just something you might need to consider if you look more into this. Chemistry is hard lol.
yes, I was coming to the same conclusion... it is difficult. What about one litre of 2M NANO2 aqueous solution? How many grams of NANO2 would that contain?
Hey, i just done a quick calculation, assuming I know what I am doing ~135g of SN are in 1 L of 2 M solution. I'll show my calculation below so you can check but I think it's pretty simple.
The unit M actually stands for mol/ L and you only have 1 L so using the equation...
n = concentration x V
You simple get
n = 2 mol/L x 1 L
n = 2
n stands for number of moles to convert number of moles to mass you need the molar mass (GFM) of your conpound, according to wikipedia for SN this id 68.9953 g/mol
Standard units are Kilograms but since you probably want the answers in grams I am just going to use grams as my unit in the equation...
n = m/GFM
2 mol = m/68.9953 g/mol
m = 137.9806 g
No idea how that lines up with lethal dosage but I know that if you get a solution in that high of a concentration you likely need to dilute it to consume it, wont change how much is in there (providing you use all the original solution) should just make it less likely to come back up.
Hope this helps at all