SOLUTION: A chemist has two solutions of HC1. One has a 40% concentration and the other has a 25% concentration. How many liters of each solutions must be mixed to obtain 129 liters of 38% s
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Question 943106: A chemist has two solutions of HC1. One has a 40% concentration and the other has a 25% concentration. How many liters of each solutions must be mixed to obtain 129 liters of 38% solution.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
A chemist has two solutions of HC1. One has a 40% concentration and the other has a 25% concentration. How many liters of each solutions must be mixed to obtain 129 liters of 38% solution.
Let x= amount of 40% concentration needed
Then 129-x=amount of 25% concentration needed
Now we know that the amount of pure HC1 that exists before the mixture takes place ((0.40x)+0.25(129-x))has to equal the amount of pure HC1 that exists after the mixture takes place(0.38*129). Sooooo:
0.40x+0.25(129-x)=0.38*129
0.40x+32.25-0.25x=49.02
0.15x=16.77
x=111.8 liters ---amount of 40% concentration needed
129-x=129-111.8=17.2 liters ----amount of 25% concentration needed
CK
0.40*111.8+0.25*17.2=0.38*129
44.72+4.3=49.02
49.02=49.02
Hope this helps-----ptaylor
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