SOLUTION: please help with this problem a 10% acid solution is to be mixed with a 50% acid solution in order to get 120 ml of a 20% acid solution. how many ml of the 10% solution and 50% sh
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Question 283738: please help with this problem a 10% acid solution is to be mixed with a 50% acid solution in order to get 120 ml of a 20% acid solution. how many ml of the 10% solution and 50% should be mixed? Found 2 solutions by oberobic, richwmiller:Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website! With solution problems, you need to keep track of the amount of 'pure' stuff.
You need to end up with 120 ml of a 20% solution, so that is
.20 * 120 = 24 ml of 'pure' acid
120-24 = 96 ml of water
.
x = volume of 10% acid
y = volume of 50% acid
.
x+y = 120 ml
.
It is better to set up the equation with the minimum number of unknowns.
In this case, we know the 120 ml is required, so we can say:
x = volume of 10% acid
120-x = volume of 50% acid
.
.1*x + .5(120-x) = .2*120
.
multiply both sides of the equation by 10 to get rid of decimals
.
10*(.1*x + .5(120-x)) = 10*(.2*120)
10*(.1*x) + 10(.5(120-x)) = 10*(.2*120)
x + 5(120-x) = 2*120
.
x + 600 -5x = 240
.
-4x = -360
.
x = 90
.
y = 120-90 = 30
.
Checking, how much 'pure' stuff we would have...
.1*90 = 9 ml
.5*30 = 15 ml
9+15 = 24 ml, which is what we needed
.
So the answer is:
90 ml of a 10% solution
plus 30 ml of a 50% solution
to produce 120 ml of 20% solution.
.
Done
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