Asked by peter
What are the normality and molarity of a solution prepared by dissolving 8.050g Ba(OH)2•8H2O in 1500ml of solution?
Answers
Answered by
DrBob222
M = mols/L = 8.050/molar mass Ba(OH)2 /1.500 L = ?
Technically, one cannot determine the equivalent weight without know the reaction/titration; however, usually when no reaction is given we count the number of H or OH in the molecule. In the case of Ba(OH)2 it would be eq. wt. = molecular weight/2, then
N = # equivalents/L so that would be
N = (8.050 g/molar mass/2)/1.500
Technically, one cannot determine the equivalent weight without know the reaction/titration; however, usually when no reaction is given we count the number of H or OH in the molecule. In the case of Ba(OH)2 it would be eq. wt. = molecular weight/2, then
N = # equivalents/L so that would be
N = (8.050 g/molar mass/2)/1.500
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.