U of S | Mailing List Archive | alt-photo-process-l | Re: Newbie in lith print

Re: Newbie in lith print



Loris,
I don't know where you found that page...
Paraformaldehyde is a polymer of formaldehyde: the former is (CH2O)n (CAS n.30525-89-4), the latter is CH2O. Hence, you can use the weight of paraformaldehyde like it were pure (100%) formaldehyde, but you will need to take into account the weight/weight percentage of the solution and its density in order to calculete the corresponding liquid amount.
As an example, if 37.5 g of paraformaldehyde are required and you have a 37% (w/w) formaldehyde solution @ 1.09 kg/dm3 (this should be clearly written in the label), you will need 37.5x100/37x1/1.09= 99 cc of solution.
Alberto
I found a page saying "...A 10% formalin solution should be equivalent
to a 4% paraformaldehyde (by weight) solution...". Concentrated /
saturated formalin is 37% AFAIK, not 40%. Anyway, I'll do the math:
Since it says 100g 4% paraformaldehyde (4g paraformaldehyde) solution is
equivalent to 100g 10% formalin (10g formaldehyde) then you have to use
2.5 units of formaldehyde per 1 unit of paraformaldehyde (10 / 4 = 2.5).
Since Christina's formula calls for 37.5g paraformaldehyde, you have to
use 37.5 x 2.5 = 93.75g formaldehyde. In 37% solution terms this should
make 93.75 / 0.37 ~= 253ml 37% formalin, *if my logic is correct*... I
don't know if the information I got from that page is correct and most
importantly I'm not sure if my logic is correct. Therefore don't rely on
this info ;)
Regards,
Loris.