Question 156684: Thanks for any insight you may be able to give me...
This question is from a quiz my instructor gave to us. He says the questions come from a textbook.
How many liters of a 70% acid solution are needed to be mixed with 18 liters of a 10% acid solution to get a mixture which is 25% alcohol?
If the solution contains acid/alcohol you would need 75% acid to leave you with 25% alcohol. How is it even possible to get 75% when both solutions you are using are diluted below 75%. As far as I have tried I see no possible solution to this problem, if he meant 25% acid then it would be easy to answer.
Let me know what you think, thanks for your time,
Bill W
Answer by mangopeeler07(462)   (Show Source):
You can put this solution on YOUR website! How many liters of a 70% acid solution are needed to be mixed with 18 liters of a 10% acid solution to get a mixture which is 25% alcohol?
Alcohol? Yes, he probably meant to say "acid". Let's work it out for acid.
How many liters of a 70% acid solution are needed to be mixed with 18 liters of a 10% acid solution to get a mixture which is 25% ACID?
(amount of liters)
coefficient=percent
70(x)+10(18)=25(x+18)
Distribute everything.
70x+180=25x+450
Subtract 25x from both sides
45x+180=450
Subtract 180 from both sides
45x=270
Divide both sides by 45
x=6
Answer: 6 liters
---------------------------------------------------
Now let's assume he actually meant alcohol. 25% alcohol means 75% acid. So:
(amount of liters)
coefficient=percent
70(x)+10(18)=75(x+18)
Do you follow?
25 is changed to 75 because it is 75% acid now.
Distribute, like before
70x+180=75x+1350
Subtract 75x from both sides
-5x+180=1350
Subtract 180 from both sides
-5x=1170
Divide by -5
x=-234
Negative means you would have to subtract, and logically speaking, you wouldn't take away acid to increase the percentage of acid. Therefore, the logical answer is:
6 liters
-------------------------------------------------------
I hope this helped!
-Alani
|
|
|
