- Download PDF
How do I solve the following problem?
A mixture of 12% chlorine is to be mixed with a second mixture containing 30% chlorine. How much of the 12% mixture is needed to mix with 80 mL of the 30% solution to make a final solution of 150 mL with a 20% chlorine concentration?
(I am confused because if we know there is 80mL of the 30% mixture and the result is 150mL, doesn't the answer have to be 70mL even if the percentages don't work out?)
1 Answer | Add Yours
You are correct that the problem as stated cannot be solved. I reworded the problem just a little bit to make it clearer. Still, however, the problem is not solvable. I believe that the author of the problem might have intended something like the following:
How much of a 12% solution of Chlorine and how much water must be added to 90 ml of a 30% chlorine solution to produce 150 ml of a 20% chlorine solution? If the problem is stated this way it is solvable as follows:
Desired solution has volume of 150 ml., including 30 ml chlorine (20% of 150) and 120 ml water.
Starting with 80 ml of a 30% solution, we have 24 ml chlorine.
You need 6 more ml of chlorine for the final solution. That would be 50 ml of the 12% solution.
You mix them and end up with 130 ml volume containing the desired 30 ml chlorine.
Add 20 ml water, and you end up with a final volume of 150 ml, containing 30 ml chlorine, or a 20% chlorine solution.
This is one possible solution to the dilemma posed by the incorrectly written question. I suggest you re-check the source of the question and the exact wording of the problem to see if it might actually be similar to what I have proposed here.
We’ve answered 320,166 questions. We can answer yours, too.Ask a question