SOLUTION: You need a 10% alcohol solution. On hand, you have a 130 mL of a 85% alcohol mixture. How much pure water will you need to add to obtain the desired solution? This is what I hav

Question 1053884: You need a 10% alcohol solution. On hand, you have a 130 mL of a 85% alcohol mixture. How much pure water will you need to add to obtain the desired solution?
This is what I have done.
130(.85)+x=130+x(.10)
110.5+x=13+x
-13 -x -13 -x
97.5=x
Found 2 solutions by Boreal, macston:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The x's disappear in your last part.
130(0.85)+x(0.0)=(130+x)(0.10). The percentage of alcohol for the x on the left is 0.
110.5=13+0.10x
97.5=0.10x
x=975 ml
That seems like a lot, but one is starting with 85% and has to dilute it about 1 in 7-8 parts, so it is not surprising more water is needed.

Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
Find the amount of alcohol you have.
x=amount of alcohol
.
0.85(130)=x
110.5=x
.
y=total 10% solution:
0.10(y)=110.5
y=110.5/0.1
y=1105
The total amount of 10% solution you can make is 1105 ml.
.
Of the 1105 ml, 130 ml is the original 85% solution.
So the amount of water to add:
1105 ml - 130 ml = 975 ml
You should add 975 ml water.
.

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