# 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?) 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.

Good luck.

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