# Practicing some dilution problems

1. Jul 27, 2006

### z-component

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.

2. Jul 27, 2006

### Bystander

Correct
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.

3. Jul 27, 2006

### z-component

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.

4. Jul 27, 2006

### Bystander

Don't do that --- "the solution" without referrant, followed by "the solution" without referrrant --- it's called "fishing," and it irritates me.

Sample = 13; dilute to 21; assay of 15 = 2.7; part one of your OP suggests you understand mass balance --- now, do the mass balance for the sample, the dilution, and the assay, and tell me where all the lead from the 13 ml sample is.

5. Jul 28, 2006