Asked by Leena
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?
Answers
Answered by
DrBob222
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.