SOLUTION: You need a 35% alcohol solution. On hand, you have a 140 mL of a 65% alcohol mixture. How much pure water will you need to add to obtain the desired solution? You will need ___

SOLUTION: You need a 35% alcohol solution. On hand, you have a 140 mL of a 65% alcohol mixture. How much pure water will you need to add to obtain the desired solution? You will need ____

Question 1162819: You need a 35% alcohol solution. On hand, you have a 140 mL of a 65% alcohol mixture. How much pure water will you need to add to obtain the desired solution?
You will need _____ mL of pure water to obtain ______mL of the desired 35% solution.
Found 3 solutions by ikleyn, MathTherapy, Alan3354:
Answer by ikleyn(52802)   (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 = 0.35. From the equation, 0.065*140 = 0.35*(140+x) 0.065*140 = 0.35*140 + 0.35x x = = 120 mL. ANSWER. 120 mL of water should be added.

Solved.

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In this site, there is entire bunch of lessons covering various types of mixture problems
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts
    - Word problems on mixtures for dry substances like candies, dried fruits
    - Word problems on mixtures for dry substances like soil and sand
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive

    - Special type mixture problems on DILUTION adding water (*)
    - Increasing concentration of an acid solution by adding pure acid
    - Draining-replacing mixture problems
    - How much water must be evaporated

    - OVERVIEW of lessons on word problems for mixtures
in this site.

A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.

The most closest to your problem is the lesson marked (*) in the list.

Read the lessons and become an expert in solution the mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Mixture problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

You need a 35% alcohol solution. On hand, you have a 140 mL of a 65% alcohol mixture. How much pure water will you need to add to obtain the desired solution?
You will need _____ mL of pure water to obtain ______mL of the desired 35% solution.

Let amount of water to add be W
Percent of water in 65% mixture: 35
Required percent of water in 35% mixture: 65
We then get following: .35(140) + W = .65(140 + W)
49 + W = 91 + .65W
.35W = 42
Amount of water to add, or
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
You will need _____ mL of pure water to obtain ______mL of the desired 35% solution.
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