SOLUTION: A Chemist has a mixture that is 10% iodine and another that is 50% iodine. How much of each solution should the druggist use to get 100 ml of a solution that is 20% iodine

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Question 531060: A Chemist has a mixture that is 10% iodine and another that is 50% iodine. How much of each solution should the druggist use to get 100 ml of a solution that is 20% iodine
Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A Chemist has a mixture that is 10% iodine and another that is 50% iodine. How much of each solution should the druggist use to get 100 ml of a solution that is 20% iodine
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Equation:
active + active = active
0.10x + 0.50(100-x) = 0.20*100
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Multiply thru by 100 to get:
10x + 50*100 - 50x = 20*100
-40x = -30*100
x = 75 ml (amt. of 20% solution needed)
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100-x = 25 ml (amt. of 50% solution)
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Cheers,
Stan H.
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Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
The chemist needs 100 ml of a 20% solution.
.2*100 ml = 20 ml of pure iodine in the 100 ml solution.
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The chemist has 10% solution and 50% solution.
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x = 10% solution
y = 50% solution
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x + y = 100 ml
x = 100-y ml
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.1x + .5y = .2*100
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multiply by 10 to eliminate the decimals
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1*x + 5*y = 2*100
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substitute
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(100-y) + 5y = 200
100 + 4y = 200
4y = 100
y = 25 ml of 50% iodine
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x = 100-25
x = 75 ml of 10% iodine
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Check the amount of pure stuff to be sure this is the answer.
.5*25 = 12.5 ml
.1*75 = 7.5 ml
12.5 + 7.5 = 20 ml, which is the amount we need.
Correct.
.
Done.