SOLUTION: A chemist mixes distilled water with a 90% solution of sulfuric acid to produce a 50% solution. If 5 liters of distilled water is used, how much 50% solution is produced?
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Question 100199This question is from textbook
: A chemist mixes distilled water with a 90% solution of sulfuric acid to produce a 50% solution. If 5 liters of distilled water is used, how much 50% solution is produced? This question is from textbook
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Let x=amount of 50% solution that is produced
Then x-5=amount of 90% solution that was mixed with the distilled water
(1) Pure sulfuric acid in the 90% solution=0.90(x-5)
(2) Pure sulfuric acid in the distilled water=0
(3) Pure sulfuric acid in the final mixture=0.50x
Now we know that (1)+(2)=(3). So our equation to solve is:
0.90(x-5)=0.50x get rid of parens
0.90x-4.5=0.50x subtract 0.90x from both sides
0.90x-0.90x-4.5=0.50x-0.90x collect like terms
-4.5=-0.40x divide both sides by -0.40
x=11.25 liters -------------amount of 50% solution produced
CK
0.90*(11.25-5)=0.50*(11.25)
0.90(6.25)=5.625
5.625=5.625
