SOLUTION: In a chemistry class, 5 liters of a 4% silver iodide solution must be mixed with 10% solution to get a 6% solution. How many liters of the 10% solution are needed?
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Question 1093510: In a chemistry class, 5 liters of a 4% silver iodide solution must be mixed with 10% solution to get a 6% solution. How many liters of the 10% solution are needed?
Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52790) (Show Source): You can put this solution on YOUR website!
.
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.
For example, your sample/template is the problem 2 of the lesson marked (*) in the list above.
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(39618) (Show Source): You can put this solution on YOUR website!
In water as the solvent or something else?
4% silver iodide in water is nonsense. Maybe at best 0.004 grams per 100 ml. water at 20C.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
I'm not a chemist, so I don't know whether or not the problem as stated makes sense in the real world. But people who write these questions aren't chemists either; their purpose is to write problems to give students practice in solving mixture problems. So ignore the possibility that the actual ingredients are nonsense and concentrate (haha - a pun) on the mathematics.
Solving this problem by the standard algebraic process for solving mixture problems, we would say
5 liters of 4% solution...
plus x liters of 10% solution...
yields (5+x) liters of 6% solution
The equation, equating the amounts of actual chemical in the two ingredients and in the final mixture, is
A bit of not-too-ugly algebra leads to the answer to the problem.
But here is a much faster way to the answer to the problem:
The 6% of the final solution is twice as close to 4% as it is to 10%.
(That is, from 4% to 6% is 2%; from 10% to 6% is 4%; the 2% difference is half of the 4% difference, which means the percentage of the final mixture is twice as close to 4% as it is to 10%.)
But twice as close to 4% as to 10% means the mixture must contain twice as much of the 4% ingredient as the 10% ingredient.
So if there were 5 liters of the 4% solution, we need half as much of the 10% solution, or 2.5 liters.
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