Posted by Anonymous on .
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 80.0 ml left. 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 3.407g. What was the concentration ofthe original lead (II) nitrate solution?
Convert 3.407g PbCl2 to mols. That will be mols Pb(NO3)2 in 2.00 mL solution. That times 80/2 will be the mols in 80 mL and mols in the original 100 mL. Calculate molarity from there.Post your work if you get stuck. We will need that to know where you are having a problem.