Asked by S.Dunn
what is the molarity of NaF in the final solution if a solution was prepared by mixing 35.0mL of 0.200M NaF with 28.5mL of 0.350M NaF
Answers
Answered by
Steve
divide total solute by total volume of solution
solute: .035*.200 + .0285*.350 = .016975
solution: .035 + .0285 = .0635
so, we have .016975 moles/.0635L = .267M
solute: .035*.200 + .0285*.350 = .016975
solution: .035 + .0285 = .0635
so, we have .016975 moles/.0635L = .267M
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