# SOLUTION: A solution containing 45% water is mixed with another solution containing 10% bromine. If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to

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 Question 622587: A solution containing 45% water is mixed with another solution containing 10% bromine. If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to make a 34.5% water/bromine solution? Answer by ankor@dixie-net.com(15624)   (Show Source): You can put this solution on YOUR website!A solution containing 45% water is mixed with another solution containing 10% bromine. If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to make a 34.5% water/bromine solution? : The 45% water solution must be a 55% bromine solution let x = amt of 55% bromine solution then 30-x = amt of 10% bromine : A typical mixture equation .55x + .10(30-x) = .345(30) .55x + 3 - .1x = 10.35 .55x - .10x = 10.35 - 3 .45x = 7.35 x = 7.35/.45 x = 16 liters of the 55% bromine solution (45% water) and 30 - 16 = 13 liters of 10% solution