this is one kind of problem that always warps my brain. i'm trying to wrap my head around this:
how many liters of a 70% alcohol solution must be added to 50 liters of a 40% alcohol solution to produce a 50% alcohol solution?
liters of solution % alcohol
70% solution x
---------------------------------------------------------------------- step one, solving for liters of solution
total liters of alcohol
70% solution x .7 .7x
40% solution 50 .4 (.4)(50) = 20
50% solution mixture 50+x (adding) .5 (.5)(50 + x)
.7x+20 = .5(50+x)
Solve for x
x= 25 liters
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how many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?
x = ounces of pure water
ounces of solution saline percentage total
x 0x 0
50 .15 .15(50)
x+50 .10 .10(x + 50)
0 + .15(50) =.10x + 5 7.5 = .10x + 5 x = 250
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