Question 1093510: In a chemistry class, 5 liters of a 4% silver iodide solution must be mixed with 10% solution to get a 6% solution. How many liters of the 10% solution are needed?
Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52879)   (Show Source):
Answer by josgarithmetic(39630)  (Show Source):
You can put this solution on YOUR website! In water as the solvent or something else?
4% silver iodide in water is nonsense. Maybe at best 0.004 grams per 100 ml. water at 20C.
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
I'm not a chemist, so I don't know whether or not the problem as stated makes sense in the real world. But people who write these questions aren't chemists either; their purpose is to write problems to give students practice in solving mixture problems. So ignore the possibility that the actual ingredients are nonsense and concentrate (haha - a pun) on the mathematics.
Solving this problem by the standard algebraic process for solving mixture problems, we would say
5 liters of 4% solution...
plus x liters of 10% solution...
yields (5+x) liters of 6% solution
The equation, equating the amounts of actual chemical in the two ingredients and in the final mixture, is
A bit of not-too-ugly algebra leads to the answer to the problem.
But here is a much faster way to the answer to the problem:
The 6% of the final solution is twice as close to 4% as it is to 10%.
(That is, from 4% to 6% is 2%; from 10% to 6% is 4%; the 2% difference is half of the 4% difference, which means the percentage of the final mixture is twice as close to 4% as it is to 10%.)
But twice as close to 4% as to 10% means the mixture must contain twice as much of the 4% ingredient as the 10% ingredient.
So if there were 5 liters of the 4% solution, we need half as much of the 10% solution, or 2.5 liters.
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