Question 277315: one acetic acid solution is 70% water and another is 30% water. how many liters of each solution should be mixed to produce 20 liters of a solution that is 40% water?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! With 'solution' problems, you have to keep track of how much pure stuff you need.
This question is tricky because the question presents the complementary percentages.
Solution 1 is 70% water, so it is 30% pure acetic acid. We normally call that a 30% solution.
Solution 2 is 30% water, so it is 70% pure acetic acid. We normally call that a 70% solution.
The goal is 20 liters of a 60% acetic acid solution.
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x = volume of 30% solution
20 - x = volume of 70% solution
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.3x = amount of pure acetic acid
.7(20-x) = the other amount of pure acetic acid.
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20 liters of 60% acetic acid will have .6*20 = 12 liters of pure acetic acid.
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.3x + .7(20-x) = 12
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.3x + 14 - .7x = 12
-.4x = -2
x = -2/-.4 = 5 liters of 30% solution
which means there is
15 liters of 70% solution.
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checking to see if we have 12 L of pure acetic acid
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.3*5 + .7*15 = 1.5 + 10.5 = 12
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So, 5 liters of 30% acetic acid + 15 liters of 70% acetic acid results in 20 liters of 60% acetic acid solution.
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Reviewing the question, we need to give the complementary answer in terms of percentage of water.
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Answer:
5 liters of 70% water + 15 liters of 30% water = 20 liters of 40% water
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