SOLUTION: If you need 425mL of a .90% saline solution but only have 25.0% saline, how would you go about making the .90% solution? Do calculation and explain in words.

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Question 508286: If you need 425mL of a .90% saline solution but only have 25.0% saline, how would you go about making the .90% solution? Do calculation and explain in words.
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
With mixture problems you have to determine how much 'pure' stuff you need.
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You need 425 mL of a 0.90% saline: So you will have 3.825 mL of 100% saline + 421.175 mL of water.
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You have only 25% saline solution. That is more than 25 times the strength you need.
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.25x = 3.825
x = 3.825/.25
x = 15.3 mL
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In 15.3 mL of 25% saline you have all the 'pure' saline you need.
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Add 425-15.3 = 409.7 mL of water.
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409.7 mL of water + 15.3 mL of 25% saline = 425 mL of 0.90% saline
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Done.