SOLUTION: a 5 liter container is full of a 28 percent acid solution which needs to be diluted to a 20 percent acid solution. How much of the 28 percent must be drained off and replaced with

Algebra.Com
Question 1095515: a 5 liter container is full of a 28 percent acid solution which needs to be diluted to a 20 percent acid solution. How much of the 28 percent must be drained off and replaced with distilled water?
I just needed help setting up the problem. The distilled water confused me as to what values I should put in.

Found 3 solutions by greenestamps, ikleyn, josgarithmetic:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Distilled water is 0% acid.

And don't let the "draining off" confuse you. In the end, you will be mixing some amount x liters of 0% acid with another amount (5-x) of 28% acid to end up with 5 liters of 20% acid.

Re-post if that is not enough to get you going.

Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
Let W be the volume to drain off from 5 liters acid solution.


Step 1:  Draining.  After draining,  you have 5-W liters of the 28% acid solution.



Step 2:  Replacing.  Then you add W liters of the distilled water (the replacing step).

                     After the replacing,  you have the same total liquid volume of 5 liters.


It contains 0.28(5-W) pure acid.

So, your "concentration equation" is


 = 0.2.    (1)


The setup is done and completed.


Now you need to solve your basic equation  (1),  and since you didn't ask me to do it, I leave it to you.


---------------
There is entire bunch of introductory 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 alloys
    - Typical word problems on mixtures from the archive
in this site.

Read them and become an expert in solution mixture word problems.
Specifically, for draining-replacing see the problems in the lesson marked (*) in the list.

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 textbook in the section "Word problems" 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 josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
This will help you.
https://www.algebra.com/my/drain-and-replace-antifreeze.lesson?content_action=show_dev
https://www.algebra.com/my/drain-and-replace-antifreeze.lesson?content_action=show_dev

RELATED QUESTIONS

a chemist needs 6 liters of 20 percent acid solution. he has available 25 percent acid... (answered by mananth)
a chemist needs 6 liters of 20 percent acid solution. he has available 25 percent acid... (answered by mananth)
a chemist mixes an 8 percent HCL acid solution with a 5 percent HCL acid solution how... (answered by josgarithmetic)
Determine how many liters of a 10 percent acid solution and 25 percent acid solution... (answered by Alan3354)
Two liters of a 3% acid solution needs to be diluted to a 2% solution. How many liters of (answered by josmiceli)
a 70% acid is to be mixed with water to produce a 5 liter mixture containing 28% acid.... (answered by josmiceli)
A container holds 10 pints of a solution which is 20% acid if 3 quarts of pure acid or... (answered by josmiceli,ikleyn)
How many quarts of water must be added to 40 quarts of solution which is 5 percent acid... (answered by lwsshak3)
Four hundred grams of an acid solution, which is 12 percent pure acid, should be diluted (answered by chessace)