Molarity and Dilution

[Chem_Mod]

A solution is prepared by dissolving 55.1 g of KCl in approximately 75 mL of water and then adding water to a final volume of 125 mL. What is the molarity of KCl(aq) in this solution?

[Chem_Mod]

molarity of KCl(aq)＝55.1g/(74.5513g/mol)/(0.125L)=5.91 mol/L (assuming 75 ml of water is an aliquot of water).

[Kevin Hernandez 3A]

Why is this problem not a Mi * Vi= Mf* Vf, if it is giving you the final volume (125ml).

[Yixin Angela Wang 2H]

Molarity is moles of solute/volume of solution. The Mi*Vi = Mf*Vf is dilution, which is not what this particular problem is asking. To solve this problem, you first convert grams of KCl to moles by dividing the mass by the molar mass. Then, you divide the number of moles by the volume in liters, which would be 0.125L. The quotient would be your molarity.

[DavidEcheverri3J]

So the 75 mL of the original solution is not required to solve the final molarity?

[Liza Hayrapetyan-3K]

No, because it mentions how the final volume BECOMES 125 mL meaning it replaces the 75 mL. I think that part was meant to test us.

[SummerLee1F]

I would use the equation Minitial x Vinital = Mfinal x Vfinal, based on the information we are given.

[Yixiao Hu 3C]

molarity=m(kcal)/M(kcal)/V(KCL)

[Ronald Thompson 1F]

If done correctly, should I have gotten 5.92?
I also was not entirely sure how to implement Mi x Vi = Mf x Vf

[Dakota Walker 1L]

I still don't understand why this isn't a dilution when a 75mL solution was created at first, then this same solution was diluted to get a 125mL solution. In what way would a dilution problem be written differently?

[Justin Quan 4I]

For this problem, you do not use M1V1=M2V2 because there’s no initial or final molarity or volume. It’s simply just moles/(total volume). The total volume being 125ml.

[Mary Koutures 2I]

The molarity of the solution is the same in the 75 mL of water and the 125 mL of water. Since the molarity is the same, we are only looking for the molarity of the final solution. We can also find the molarity of the first solution by doing M=.74 Moles/.075L= 9.87 mol/L. The molarity of the first solution isn't necessary in this exercise, but it helps us conceptualize the fact that with the same amount of moles and less water, the concentration is higher than the second solution.