SOLUTION: A chemist needs 200 milimeters of a 49% acid solution but only has 37% and 57% acid solutions available. Find how many milimeters of each that should ne mixed to get the desired s
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-> SOLUTION: A chemist needs 200 milimeters of a 49% acid solution but only has 37% and 57% acid solutions available. Find how many milimeters of each that should ne mixed to get the desired s
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Question 221269: A chemist needs 200 milimeters of a 49% acid solution but only has 37% and 57% acid solutions available. Find how many milimeters of each that should ne mixed to get the desired solution. Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! .57x+.37(200-x)=.49*200
.57x+74-.37x=98
.20x=98-74
.20x=24
x=24/.20
x=120 ml. of th 57% mixture is used.
200-120=80 ml of the 37% mixture is used.
Proof:
.57*120+.37*80=.49*200
68.4+29.6=98
98=98
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