# Practicing some dilution problems

I've been practicing some dilution problems from my chemistry book and I can't seem to get some of these down.

A 13 mL sample of a solution of Pb(ClO3)2 was diluted with water to 21 mL. A 15 mL sample of the dilute solution was found to contain 2.7 moles of Pb2+. What was the concentration of Pb(ClO3)2 in the original undiluted solution?

The concentrated solution has a 13 mL sample with an unknown molarity.
The diluted solution has a 15 mL sample with an unknown molarity and a given amount of moles for the Pb ion.

Since the ratio of Pb ions to Pb(ClO3)2 is 1:1, there are also 2.7 moles of Pb(ClO3)2 (I think). I divided 2.7 mol Pb by .015 L (15 mL) and got a molarity of 180 M.

Now since the amount of moles from the concentrated to the diluted solutions has to be the same, I divided 2.7 by .013 L to get 207 M. The correct answer is 290 M. I'm not sure what I did wrong.

It would be much appreciated if someone could point me in the right direction.

## Answers and Replies

Bystander
Homework Helper
Gold Member
z-component said:
(snip)Now since the amount of moles from the concentrated to the diluted solutions has to be the same,

Correct
I divided 2.7 by .013 L

Why 2.7 by 0.013? It's wrong, and if you can tell me why you did it, you'll know why it's wrong.

Is it correct to say that 2.7 moles of the solution allow me to directly find the molarity of the solution if I know the volume? That is what I did. Perhaps 2.7 isn't the measure for the entire solution but only the solute.

Bystander