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16 Sep 2014, 00:42
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
44% (02:12) correct
56%
(01:56)
wrong
based on 389
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A strong acid solution weighing 80 grams was diluted by adding \(x\) grams of water. After dilution, the concentration of the acid in the solution became \(\frac{1}{y}\) times its initial concentration. What was the concentration of the acid in the original solution?
(1) \(x = 80\)
(2) \(y = 2\)
(1) \(x = 80\)
(2) \(y = 2\)
16 Sep 2014, 00:42
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Official Solution:
A strong acid solution weighing 80 grams was diluted by adding \(x\) grams of water. After dilution, the concentration of the acid in the solution became \(\frac{1}{y}\) times its initial concentration. What was the concentration of the acid in the original solution?
The question looks intimidating and one might jump right into constructing equation. But take a moment, step back and use logic.
(1) \(x = 80\).
We can see that the initial solution was diluted with an equal amount of water (80 grams of solution was diluted with 80 grams of water). Whatever the concentration of the acid was in the original solution, this operation, would halved it in the resulting solution. For example, if the initial concentration of acid was 10%, so if there were 8 grams of acid in the original solution, the concentration of acid in the resulting solution would be 5% (8 grams of acid over 160 grams). However, while we can infer that the concentration of acid was halved, we don't have any information on what it actually was. Therefore, statement (1) alone is not sufficient to determine the original concentration of acid.
(2) \(y = 2\)
This statement also tells us that the concentration of acid was halved in the resulting solution, but we have no idea what the original concentration actually was. Therefore, statement (2) is also not sufficient.
(1)+(2) Notice that both statements convey the same information: the concentration of the acid in the original solution was halved in the resulting solution, but we still lack information about its actual value. Therefore, combining the statements is not sufficient to determine the original concentration of the acid.
Answer: E
A strong acid solution weighing 80 grams was diluted by adding \(x\) grams of water. After dilution, the concentration of the acid in the solution became \(\frac{1}{y}\) times its initial concentration. What was the concentration of the acid in the original solution?
The question looks intimidating and one might jump right into constructing equation. But take a moment, step back and use logic.
(1) \(x = 80\).
We can see that the initial solution was diluted with an equal amount of water (80 grams of solution was diluted with 80 grams of water). Whatever the concentration of the acid was in the original solution, this operation, would halved it in the resulting solution. For example, if the initial concentration of acid was 10%, so if there were 8 grams of acid in the original solution, the concentration of acid in the resulting solution would be 5% (8 grams of acid over 160 grams). However, while we can infer that the concentration of acid was halved, we don't have any information on what it actually was. Therefore, statement (1) alone is not sufficient to determine the original concentration of acid.
(2) \(y = 2\)
This statement also tells us that the concentration of acid was halved in the resulting solution, but we have no idea what the original concentration actually was. Therefore, statement (2) is also not sufficient.
(1)+(2) Notice that both statements convey the same information: the concentration of the acid in the original solution was halved in the resulting solution, but we still lack information about its actual value. Therefore, combining the statements is not sufficient to determine the original concentration of the acid.
Answer: E
General Discussion
28 Nov 2014, 07:26
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let k grams of solute be present.
then,
equating final concentrations:
(1/y)*(k/80)=k/(x+80)
so , k cancel out,
we need to calculate k/80
hence ans is E.
then,
equating final concentrations:
(1/y)*(k/80)=k/(x+80)
so , k cancel out,
we need to calculate k/80
hence ans is E.
