You add 5g of PbCl2 to a solution of PbCl2. 2g of PbCl2 is undissolved as a precipitate at the bottom. What was the original solution?

Answer 1

You have either misstated the problem or else it basically doesn't make much sense. The "original solution" didn't really exist (unless you're calling the solvent, water, a solution). So, looking at it another way, if you had 5 g of PbCl2 added to a solution OF PbCl2 and 2 g of PbCl2 is undissolved, then apparently 3 grams remain in solution as Pb²⁺ and Cl^- ions.