SOLUTION: Hi, I need help solving this problem. I have a big! problem with mixture problems : and I have to take chemistry soon. Please help!! Thanks Question: A radiator contains 6L of a

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Question 67890: Hi, I need help solving this problem. I have a big! problem with mixture problems : and I have to take chemistry soon. Please help!! Thanks
Question:
A radiator contains 6L of a 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 50% antifreeze solution?
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A radiator contains 6L of a 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 50% antifreeze solution?
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Amt of 25% solution = 6 liters; amt of active ingredient= 0.25*6= 1.5 liters
Amt of 25% removed is x liters; amt of active ingre removed is 0.25x liters
Amt replaced is x liters; amt of active ingredient added = x liters
Resulting amount is 6 liters ; amt of active ingredient = 0.50(6)= 3 liters
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EQUATION:
active -active + active = active
1.5 - 0.25x + x = 3
0.75x=1.5
x=2 liters
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2 liters of the 25% solution should be drained out and replaced
with 2 liters of 100% antifreeze.
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Cheers
Stan H.