SOLUTION: For an experiment, a student requires 200 milliliters of a solution that is 8% NaCl (sodium chloride). The storeroom has only solutions that are 15% NaCl and 5% NaCl. How many mill

Question 898491: For an experiment, a student requires 200 milliliters of a solution that is 8% NaCl (sodium chloride). The storeroom has only solutions that are 15% NaCl and 5% NaCl. How many milliliters of each available solution should be mixed to get 200 milliliters of 8% NaCl?
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Do it like this one. Only the numbers are different.
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Amanda wants to make 8 gal. of a 20% saline solution by mixing together a 56% saline solution and a 8% saline solution. How much of each solution must she use?
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e = amount of 8%
f = amount of 56%
---
e+ f = 8 (total solution)
8e + 56f = 20*8 (total saline)
---
e+ f = 8
e + 7f = 20
------------ Subtract
-6f = -12
f = 2
e = 6
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
For an experiment, a student requires 200 milliliters of a solution that is 8% NaCl (sodium chloride). The storeroom has only solutions that are 15% NaCl and 5% NaCl. How many milliliters of each available solution should be mixed to get 200 milliliters of 8% NaCl?
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Equation:
salt + salt = salt
0.15x + 0.05(200-x) = 0.08*200
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15x + 5*200 - 5x = 8*200
10x = 3*200
x = 60 ml (amt. of 15% solution needed)
200-x = 140 ml (amt. of 5% solution needed)
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Cheers,
Stan H.
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