SOLUTION: What quantity x of a 75% acid solution must be mixed with a 25% solution to produce 500 mL of a 40% solution? Amount x=

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Question 260699: What quantity x of a 75% acid solution must be mixed with a 25% solution to produce 500 mL of a 40% solution?
Amount x=
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
This is a mixture problem. Here is a chart based on the above information:
acids . . . . . . . . % . . . . . . . . . .ml . . . . . . . . . . . . . %ml
75 . . . . . . . . . .75 . . . . . . . . .x . . . . . . . . . . . . . . . .75x
25 . . . . . . . . . .25 . . . . . . . . .500-x . . . . . . . . . . . 12500 - 25x
mixture . . . . . .40 . . . . . . . . .500 . . . . . . . . . . . . . .20000
using column three, we get
75x + 12500 - 25x = 20000
combine like terms to get
50x + 12500 = 20000
subtract 12500 to get
50x = 7500
divide to get
x = 150
we need 150 ml of 75% acid.