SOLUTION: How many parts of glacial acetic acid (99.5% acetic acid) must be added to 100 parts of a 10% solution of acetic acid to give a 28% solution? (Round your answer to the nearest whol

Question 71265This question is from textbook introductory algebra
: How many parts of glacial acetic acid (99.5% acetic acid) must be added to 100 parts of a 10% solution of acetic acid to give a 28% solution? (Round your answer to the nearest whole part) This question is from textbook introductory algebra

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of the 99.5% solution to be added
Now we know that the amount of pure acetic acid in the original solution (0.1(100)) plus the amount of pure acetic acid in the solution that was added (0.995(x)) must equal the amount of pure acetic acid in the final mixture ((100+x)(0.28)). So our equation to solve is:
0.1(100)+0.995(x)=(100+x)(0.28) get rid of parens
10+0.995x=28+0.28x subtract 10 and also 0.28x from both sides
10-10+0.995x-0.28x=28-10+0.28x-0.28x collect like terms
0.715x=18 divide both sides by 0.715
x=25.175 or 25 parts of the 99.5% solution must be added----------------ans
CK
0.1(100)+0.995(25)=125(0.28)
10+24.875=35
~35=35

Hope this helps------ptaylor

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