Question

100 mL of LiNO3, 0.241 M is mixed with 240 mL of 0.618 M Ca(NO3)2. What is the final concentration of NO3- in the solution?

Answer 1

Just remember the definition of molarity. It is # mols/Liter of solution. That equation can be rearranged to give #mols = M x L.
# mols LiNO3 = 0.241 M x 0.100 L = ??
# mols NO3^- from LiNO3 is the same.
# mols Ca(NO3)2 = 0.618 M x 0.240 L = ??
# mols NO3^- from Ca(NO3)2 is twice that.
Total mols NO3^- = mols from LiNO3 + mols from Ca(NO3)2 = ??
Then molarity NO3^- = mols NO3^-/L of solution.
The volume, of course, is 0.100 L + 0.240 L = ??

Answer 2

0.508M

Answer 3

0.943

Answer 4

.508