.
A flask contains 500 ml of an 80% alcohol solution in water. How much of this solution must be drawn off and replaced by water
to obtain a solution having a concentration of 25% alcohol?
~~~~~~~~~~~~~~~~
Let V be the volume to drain off from 500 mL of the original mixture.
Step 1: Draining. After draining, you have 500-V mL of the 80% mixture.
Step 2: Replacing. Then you add V mL of water (the replacing step).
After the replacing, you have the same total liquid volume of 500 mL.
It contains 0.8*(500-V) of the pure alcohol.
So, your "concentration equation" is
= 0.25. (1)
The setup is done and completed.
Now you need to solve your basic equation (1). Multiply both sides by 500.
0.8*(500-V) = 0.25*500
0.8*500 - 0.8V = 125
400 - 125 = 0.8V
V = = 343.75 mL.
Answer. 343.75 mL of the original mixture should be drained and replaced with water.
Solved.
--------------
It is a standard and typical mixture problem.
In this site, there is a bunch of lessons, covering various types of mixture problems. See introductory lessons
- 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
Of these lessons, the closest solutions to your problem you will find in 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.