You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 84.8 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 1.650 g. What was the concentration of the original lead(II) nitrate solution?

Answer 1

mols PbCl2 = grams/molar mass = approx 6E-3 but you need a better number than that. I've estimated as well as all of the calculations that follow.
That's mols in 2.00 mL out of 84.8 mL. Convert to mols in 100 mL (the original volume) by 6E-3 x (100/84.8) = approx 7E-3 mols in the original solution.
Then M = 7E-3/0.100 L = about 0.07M