Quandaries and Queries


A chemist has one solution that is 35% pure acid and another solution that is 75% pure acid. How many cubic centimeters of each solution must be used to produce 80 cubic centimeters of solution that is 50% acid? Can you please explain all the different ways of setting this up(all the different approaches to solve this type of example). I'm very good at solving once I have an equation. And I know the answer is 50 of 35% solution, 30 of 75% solution because my book has answers, but I can't figure out any equations to solve this with. You are so helpful!



You need a beaker with 80 cc's of a 50% solution so you need this beaker to contain 40 cc's of pure acid and 40 cc's of water. You don't know how much of the 35% solution to put into this beaker so give it variable value. Let X be the number of cc's of the 35% solution you put into the beaker. Since 35% of this is pure acid, you are putting 35% of x cc's of pure acid into the beaker.

You have now put x cc's into the beaker so you need (80 - x) cc's more from the 75% solution. This adds 75% of (80 - x) cc's of pure acid to the beaker. In total you want 40 cc's of pure acid, so

(35% of x) + 75% of (80 - x) = 40

that is

 35/100 x + 75/100 (80 - x) = 40

Solve for x.




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