When working with solutions, we need some way to figure out the amount of acid in the water in the original statement. if we know the total is 80 grams of solution, the possible % of acid of that solution is infinite. If you know that you add 80g water to 80g solution, this doesn't provide all the variables we need to know in order to determine the concentration of the original. We would need to know what % concentration the solution became after the 80g water is added. We have 3 variables, the 80g water added, the concentration of the original solution, and the % mixture [in the form of a fraction] after the water is added to the original solution. Knowing any 2 of the 3 can answer the question for you. Since after statement 1, we still only know 1 of the 3 variables, we do not have enough information, therefore INSUFFICIENT.
The reason this is INSUFFICIENT is the same as why Statement 1 is insufficient. We need to know 2 of the 3 variables and only know 1 [Remember that we do not know how much water is added. We CANNOT keep information from Statement 1 when considering Statement 2].
INSUFFICIENT. If we have 80g to start and we add 80 g of water, then no matter what the original concentration is, we're going to be cutting the concentration in half.
40g acid, 40g water = 40g acid / 80 total = 50% original concentration. If we add 80g water to this, we have 40g acid / 160g total = 25%, which is 1/2 the original concentration.
If you have:
60g acid, 20g water = 60/80 = 75% solution. If we add 80g water, then we get 60g acid / 160g total = 37.5%, which is 1/2 of 75%, the original concentration.
When the end % concentration is relative to the original concentration, it does not give us enough information because there are too many variables that are not nailed down, and any combination of acid-water mix, when we double the volume by adding water, this will ALWAYS cut the original concentration in half.
Xgrams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became 1/y times of the initial concentration, what was the concentration of acid in the original solution?
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
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