posted by S.Dunn on .
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
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