SOLUTION: I'm not sure how to solve this or where to even begin.
A chemist wants to dilute her 5L stock of 30% solution of acid to 15%. How much 10% solution does she have to add to the
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Question 783407: I'm not sure how to solve this or where to even begin.
A chemist wants to dilute her 5L stock of 30% solution of acid to 15%. How much 10% solution does she have to add to the 30% solution in order to obtain a mixture that is 15% acid?
Found 2 solutions by josmiceli, solver91311:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = liters of 10% solution to be added
= liters of acid in 10% solution
liters of acid already in 30% solution
------------------
15 liters of 10% solution need to be added
check:
OK
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The amount of pure acid in 5 liters of 30% solution is )
liters. The amount of pure acid in 
liters of 10% solution is 
liters. Mixed together, the amount of pure acid in 
liters of 15% solution is either \ +\ 0.1x\ )
or )
, so:
\ +\ 0.1x\ =\ 0.15(5\ +\ x))
Solve for
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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