SOLUTION: How much water must be added to a bottle of 840cm cubes of acid which contains 10% of pure acid to ontain a solution containing 7% of pure acid? Thank u

Question 1110771: How much water must be added to a bottle of 840cm cubes of acid which contains 10% of pure acid to ontain a solution containing 7% of pure acid?
Thank u
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52817)   (Show Source): You can put this solution on YOUR website!
.

Your equation is = 0.07. It says that the ratio of the acid content to the total liquid volume (after adding x cm^3 of water) is equal to the given concentration 0.07 = 7%. To solve the equation, multiply both sides by (840+x). It is your first step.

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It is a typical mixture problem.

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
    - Advanced mixture problems
    - Advanced mixture problem for three alloys
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.

Read them and become an expert in solution 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 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 greenestamps(13203)   (Show Source): You can put this solution on YOUR website!

Since what is being added is pure water, I would use mental arithmetic to solve this problem.

(1) The amount of acid is 10% of 840 cm^3, or 84 cm^3.
(2) If that same 84 cm^3 of acid is to be 7% of the final mixture, then the total volume of the final mixture must be 84/.07 = 1200 cm^3.
(3) You started with 840 cm^3 of solution and finished with 1200 cm^3; the amount of pure water you added was 1200=840 = 360 cm^3.
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