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?
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
