Take a liter of solution.
1.29 g/mL x 1000 mL = 1290 g.
Part is water and part is KOH. The KOH is 30%; therefore,
1290 x 0.3 = ?? grams KOH.
Determine mols KOH and that is mols/L and that is molarity.
Determine the molarity of 30% (w/w) KOH of specific gravity 1.29. Thanks for any help.
1 answer
1 answer