No it's not homework. I'm checking the calcs in a report by a colleague in nthe alkaline hydrolysis business and his stated 0.78 M KOH seems off to me. I get 297.01 moles in 561.86 liters to be 0.52 molarity
I'm trying to verify the final molarity of my solution.
I'm adding 40.69 lb of 90% anhydrous KOH to 148.43 (561.86L) gallons of water . I get a final molarity of 0.52. The report I'm reading arrives at a molarity of 0.78. Is this an error?
Thanks to anyone who can provide the calcs.
Hoping to get my assumptions and math verified here.
We have a process that requires 42 lbs of dry flake KOH (90% purity). Looking to find the equivalent volume (liters) of 45% liquid KOH (11.1 molar strength) - being extracted from a 55 gallon drum.
1 liter of liquid KOH has 11.1 x 56...