Asked by DIANA
What concentration of Cl– results when 993 mL of 0.521 M LiCl is mixed with 883 mL of 0.309 M MgCl2?
PLEASE HELP!
I don't know how to set this up
PLEASE HELP!
I don't know how to set this up
Answers
Answered by
DrBob222
Technically this can't be set up BECAUSE volumes are not additive; however, I'm sure the author of the problem meant to say to assume volumes were additive.
mols LiCl = M x L = ?
mols Cl^- from LiCl = same ?
mols MgCl2 = M x L = ?
mols Cl^- from MgCl2 = 2x that number.
Total mols Cl^- = sum of Cl^- from above.
Then M = mols/L solution.
mols LiCl = M x L = ?
mols Cl^- from LiCl = same ?
mols MgCl2 = M x L = ?
mols Cl^- from MgCl2 = 2x that number.
Total mols Cl^- = sum of Cl^- from above.
Then M = mols/L solution.
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