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A chemist wishes to dilute a 90-cc solution that is 5% to that is 3% acid. How much water should the chemist add to the acid? 

 Oct 23, 2018
 #1
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Let  pure water  = 0%  acid

Let the amount of water to be added  = W....so we have

 

(90 cc) * (5% acid  solution)  +  W *( 0%  acid)  = ( Final amt) * (3% acid)

 

(90) (.05)  +  W (0) =  ( 90 + W) (.03)      simplify

 

4.5   =   2.7 +  .03W      subtract 2.7 from both sides

 

1.8  = .03W    divide both sides by .03

 

1.8 / .03  =  W   =  60 cc   of water 

 

cool cool cool   

 Oct 24, 2018

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