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

1 answer

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.