Question 272568: in a chemistry class, a 4% silver iodide solution must be mixed with a 10% solution to get 12 liters of a 6% solution. how many liters of the 10% solution are needed?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! Solution problems are solved most easily by keeping track of the amount of 'pure' stuff in the constituent solutions and in the final solution.
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Here you want 12 liters of a 6% solution of silver iodide.
That means you need .06*12 = .72 liters of 'pure' silver iodide.
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You have a 4% and a 10% solution to add together.
'x' can be used to indicate the amount of 4% solution.
We could say 'y' is the amount of 10% solution, but it is likely easier to indicate '12-x' as the amount of 10% solution.
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We can now setup the equation to be solved:
.04x + .10y = .72
or
.04x + .10(12-x) = .72
.04x + 1.2 - .1x = .72
-.06x = -.48
x = 8
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y = 12-8 = 4
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So, our candidates are:
x = 8 liters of 4% solution
y = 4 liters of 10% solution
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But we have to check to know if that's right...always...
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How much silver iodide is there in 8 liters of 4% solution?
.04*8 = .32 liters
How much silver iodide is there in 4 liters of 10% solution?
.1*4 = .4 liters
Total = .72 liters of silver iodide
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That matches what we determined we needed initially.
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Checking the question, we can see the answer is only "how many liters of the 10% solution"? 4 liters
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