what is the molarity of the final solution?

Question

If 130. mL of 1.10 M KCl are mixed with 120. mL of 3.10 M KCl, what is the molarity of the final solution? Assume the volumes are additive.

Also, what are some good sites to learn solution stoichiometry because my textbook is horrible?

Answer 1

The moles of KCl in the final solution are going to be equal to the moles in the first solution plus the moles in the second solution.
Since mole=M*V you have

Mfinal*Vfinal = M1*V1+ M2*V2

but Vfinal= V1+ V2
So
Mfinal*(V1+ V2) = M1*V1+ M2*V2 =>
=> Mfinal= (M1*V1+ M2*V2)/(V1+V2)=
=(1.10*130+ 3.10*120) /(130+120)
=2.06 M

note that I didn't have to convert ml in liters since the conversion factor is simpified in the ratio.

Answer 2

Calculate the moles of KCl in each solution, then add them up. Add up the volumes. Divide the total number of moles of KCl by the total volume. Voila. Make sure your units cancel properly.

Volume x Molarity = Moles

Moles / Volume = Molarity

Answer 3

Moles HNO3 in the 1sr answer = 2.60 x 4.30 = eleven.18 Moles HNO3 in the 2d answer = one million.50 x one million.eighty = 2.70 comprehensive moles = eleven.18 + 2.70 = thirteen.88 comprehensive quantity = 2.60 + one million.50 = 4.10 L M = thirteen.88 mol / 4.10 L = 3.39