Question 1078560: How much water should be added to 6 liters of 30% acid solution to dilute it to a 12%soution
Found 3 solutions by Alan3354, Boreal, natolino_2017:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! How much water should be added to 6 liters of 30% acid solution to dilute it to a 12%soution
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Add W liters.
6*30 + 0*W = (6+W)*12
Solve for W
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! x=amount of water in liters
6(.30)+x(0)=(6+x)(.12), pure acid amount
1.8=0.72+.12x
.12x=1.08
x=9 liters
The total solution will be 15 liters
going from 6 to 15 is an increase of 2.5
the reciprocal is 0.4, and that times 30%=12% check.
Answer by natolino_2017(77)  (Show Source):
You can put this solution on YOUR website! Let W:Water on the initial solution
A:Acid on the initial solutiom.
X:Water need to be added.
1) A+W=6
2) A/(A+W)=30%
3) A/(A+W+X)=12%
Using 1) and 2)
A/6=30%, solving for A, A=1,8 liters.
So in 3) 1,8/(6+x)=12%, solving for x, 6+x=15, x=9 liters.
Answer: you must add 9 liters of water in order to make the solution of 12%.
@natolino_
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