SOLUTION: hector is performing a chemistry experiment that requires 140 milliliters of a 30% copper sulfate solution. He has a 25% copper sulfate solution and a 60% copper sulfate solution.

Question 104828: hector is performing a chemistry experiment that requires 140 milliliters of a 30% copper sulfate solution. He has a 25% copper sulfate solution and a 60% copper sulfate solution. How many milliliters of each solution should he mix to obtain the needed solution?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

Let x=amount of 25% copper sulfate solution needed
Then 140-x=amount of 60% copper sulfate solution needed
Now we know that the amount of pure copper sulfate in the 25% solution (0.25x) plus the amount of pure copper sulfate in the 60% solution (0.60(140-x)) has to equal the amount of pure copper sulfate in the final mixture(0.30*140). So our equation to solve is:
0.25x+0.60(140-x)=).30*140 get rid of parens
0.25x+84-0.60x=42 subtract 84 from both sides
0.25x+84-84-0.60x=42-84 collect like terms
-0.35x=--42 divide both sides by -0.35
x=120 ml ---------------------------------- amount of 25% solution
140-x=140-120=20 ml------------------------------amount of 60% solution
ck
0.25*120+0.60(140-120)=0.30*140 simplify
30+12=42
42=42
Hope this helps---ptaylor

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