Asked by Gelo
                Pure water is being added to a 25% solution of 120 milliliters of hydrochloric acid. How much water should be added to reduce it to a 15% mixture ?
            
            
        Answers
                    Answered by
            Steve
            
    here's one way. You want the concentration to be reduced by a factor of 15/25 = 3/5
So, the volume must be increased by a factor of 5/3.
5/3 * 120 = 200 ml
So, if you add 80ml of water, the concentration is reduced from 25% to 15%
Or, you can solve the equation
.25*120 + 0x = .15(120+x)
and arrive at the same answer.
    
So, the volume must be increased by a factor of 5/3.
5/3 * 120 = 200 ml
So, if you add 80ml of water, the concentration is reduced from 25% to 15%
Or, you can solve the equation
.25*120 + 0x = .15(120+x)
and arrive at the same answer.
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