SOLUTION: I have been struggling with this word problem and I don't know where to start: What amount of a 50% acid solution must be mixed with a 25% solution to produce 500 mL of a 45% solut
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Question 990833: I have been struggling with this word problem and I don't know where to start: What amount of a 50% acid solution must be mixed with a 25% solution to produce 500 mL of a 45% solution?
If you could just help me get started that would be greatly appreciated. Found 2 solutions by Alan3354, ikleyn:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Do it like this one:
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One alloy contains 80% gold (by weight)and another alloy contains 55% gold.how many grams of each alloy should be combined to make 40g of an alloy that contains 70% gold.
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e = amount of 80%
f = amount of 55%
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e + f = 40 (total alloy)
80e + 55f = 70*40 (total gold)
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You can put this solution on YOUR website! .
What amount of a 50% acid solution must be mixed with a 25% solution to produce 500 mL of a 45% solution?
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Let x be the volume of a 50% acid solution (in mL).
Then the volume of the pure acid in this solution is 0.5x mL.
Next, the volume of a 25% solution is (500-x) mL, and it contains 0.25*(500-x) of the pure acid.
Now we have 0.5x + 0.25*(500-x) of the pure acid in the mixt, and it is 45% of 500 mL, i.e 0.45*500 = 225 mL.
Thus we have an equation
0.5x + 0.25(500-x) = 225.
Solve it:
0.5x + 125 - 0.25x = 225,
0.25x = 225 - 125 = 100,
x = = 400.
Answer. 400 mL of 50% acid solution should be mixed with (500-400) = 100 mL of 25% acid solution.
Check: 0.5*400 + 0.25*100 = 225 mL of pure acid.
= 0.45.
The answer is correct.
You may want to look into the lesson Mixture problems in this site where many of similar problems are considered and solved.
