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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.