SOLUTION: a chemist need 80 millitiltes of 25% solution but only has 22% and 30% solutions available How many milliliters of each that should be mixed to get the desired solution number of m

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Question 285835: a chemist need 80 millitiltes of 25% solution but only has 22% and 30% solutions available How many milliliters of each that should be mixed to get the desired solution number of mill of 22% and the number of mill for 30%
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A chemist need 80 millitiltes of 25% solution but only has 22% and 30% solutions available.
How many milliliters of each that should be mixed to get the desired solution number of mill of 22% and the number of mill for 30%.
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Equation:
active + active = active
0.22x + 0.30(80-x) = 0.25(80)
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Multiply thru by 100 to get:
22x + 30*80 - 30x = 25*80
-8x = -5*80
x = 50 milliliters (amt. of 22% solution needed in the mixture)
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80-x= 30 milliliters (amt. of 30% solution needed in the mixture)
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Cheers,
Stan H.