SOLUTION: I cant figure this one out, please help. A chemist has 60g of solution that is 70% acid. how much water should be added to make a solution that is 40% acid. Thanks!

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Question 102401: I cant figure this one out, please help. A chemist has 60g of solution that is 70% acid. how much water should be added to make a solution that is 40% acid. Thanks!
Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website!
With mixture problems, you always should try to figure out how much "pure stuff" you have to begin with and how much you will need at the end.
The chemist has 60 grams of solution that is 70% acid, to .7*60 = 42g of pure acid. Water is considered pure water with no acid.
The final solution will be 60 grams of the original solution + x grams of pure water. It will contain 42 grams of acid (all of which came from the original solution).
We want this final solution to be 40% pure acid. We can represent this as . That simplifies to . Multiply by 10 to remove the decimal points, so we now have:
Subtracting 240 from both sides, . Dividing both sides by 4, we have . So you need to add 45 g of pure water to the original solution.
ALWAYS check your answer! 60g + 45g = 105g of solution in total. It contains 42 grams of pure acid. So the soluton is . Check.