SOLUTION: You need 665 mL of a 65% alcohol solution. On hand, you have a 95% alcohol mixture. How much of the 95% alcohol mixture and pure water will you need to obtain the desired solution?

Algebra ->  Inverses -> SOLUTION: You need 665 mL of a 65% alcohol solution. On hand, you have a 95% alcohol mixture. How much of the 95% alcohol mixture and pure water will you need to obtain the desired solution?      Log On


   

Question 1162826: You need 665 mL of a 65% alcohol solution. On hand, you have a 95% alcohol mixture. How much of the 95% alcohol mixture and pure water will you need to obtain the desired solution?
You will need
? mL of pure water
and
? mL of the 95% solution.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the volume of the pure water under the question, in mL.

Then the volume of the pure alcohol in 140 mL of the 65% alcohol solution is 0.65*140 mL.

The total volume after adding x mL of water is (140+x) mL.

The concentration equation for the resulting mixture after adding water is

%280.65%2A140%29%2F%28140%2Bx%29 = 0.35.

From the equation,

0.065*140 = 0.35*(140+x)
0.065*140 = 0.35*140 + 0.35x
x = %28065%2A140+-+0.35%2A140%29%2F0.35 = 120 mL.

ANSWER. 120 mL of water should be added.

-------------
Same problem, different numbers.