SOLUTION: In a chemistry class,8 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed? How to solve th

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Question 345424: In a chemistry class,8 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed? How to solve this problem?
Found 2 solutions by Fombitz, haileytucki:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let A be the amount of 10% solution.
Write the concentration equation.
4%288%29%2B10A=6%288%2BA%29
32%2B10A=48%2B6A
4A=16
highlight%28A=4%29 liters

Answer by haileytucki(390) About Me  (Show Source):
You can put this solution on YOUR website!
(salt I end up with)/(total liters of solution) = 4%
8 milliliters of 6% saline solution has .06*8=.48 ml salt
Let w = ml of water to be added
so....
(0.48)/(8+w)=0.04
Since the variable is in the denominator on the left-hand side of the equation, this can be solved as a ratio. For example, (A)/(B)=C is equivalent to (A)/(C)=B.
(0.48)/(0.04)=(w+8)
Since w is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
(w+8)=(0.48)/(0.04)
Remove the parentheses around the expression w+8.
w+8=(0.48)/(0.04)
Reduce the expression (0.48)/(0.04) by removing a factor of from the numerator and denominator.
w+8=12
Since 8 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 8 from both sides.
w=-8+12
Add 12 to -8 to get 4.
w=4 liters