Question 119381:
How much pure alcohol must be added to 40 oz of a 25%
solution to produce a mixture that is 40% alcohol? Found 2 solutions by stanbon, checkley71:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! How much pure alcohol must be added to 40 oz of a 25%
solution to produce a mixture that is 40% alcohol?
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25% solution DATA:
Amt = 40 oz ; amt of active ingredient = 0.25*40 = 10 oz
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Pure solution DATA:
Amt = x oz ; amt of active ingredient = 100%*x = x oz
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40% Mixture DATA:
amt. = 40+x oz ; amt of active ingredient = 0.40(40+x)=16+0.4x oz
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EQUATION:
active + active = active
10 + x = 16 + 0.4x
0.6x = 6
x = 10 oz (amt. of 100% alcohol needed for the mixture)
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Cheers,
Stan H.
You can put this solution on YOUR website! .25*40+x=.4(x+40)
10+x=.4x+16
x-.4x=16-10
.6x=6
x=6/.6
x=10 oz of pure alcohol is needed.
proof
.25*40+10=.4*50
10+10=20
20=20
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