Question 521222: You need 630 mL of a 30% alcohol solution. On hand, you have a 70% alcohol mixture. How much of the 70% alcohol mixture and pure water will you need to obtain the desired solution?
You will need
mL of pure water
and
mL of the 70% solution.
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! With solution or mixture problems you need to keep track of the 'pure' stuff. In this case, they mention 'pure' water, which is not the 'pure' stuff you're interested in. You're interested in the 'pure' alcohol.
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You need 630 mL of a 30% alcohol solution, so you will have .3*630 = 189 mL of pure alcohol and 630-189 = 441 mL of water.
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However, you do not have pure alcohol to add to water. You have a 70% alcohol solution that you have to mix with additional water until it is 30%.
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.7x = 189
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x = 189/.7 = 270 mL
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So, 270 mL of a 70% alcohol solution has 189 mL of pure alcohol and 81 mL of water. That's all the pure alcohol you need to simply add 630-270 = 360 mL of water to produce the 630 mL you need.
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Answer: Add 270 mL of 70% alcohol and 360 mL of water to produce 630 mL of 30% alcohol.
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Done.
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