Zinc iodide is dissolved in 2.5 L of water to make a solution. 1.00 L of this solution is then diluted to make 12.0 L of a 0.50 M solution. What mass of zinc was added to make the original solution?

Answer 1

mols ZnI2 in final solution is
mols ZnI2 = M x L = 0.50 x 12.0L = 6.0 mols. That also = mols in the 1.00 L taken for dilution to the 12.0 L.
This 6.0 mols in the 1.00 L taken for dilution came from 2.5 L of the original; therefore, the original solution must have had 6.0 x 2.5/1.0 = 15 mols ZnI2 to start. g ZnI2 = mols ZnI2 x molar mass ZnI2 = ?g in the original.

Check: grams ZnI2/molar mass ZnI2 = 15.0 mols. Place that in 2.5 L so you have 15.0 mols/2.5L = 6 molar solution.
You take 1.00 L of this and dilute to 12.0 L. The final concentration is
6.0 M x (1.0 L/12.0) = 0.50 M and that's the final concentration in the problem.