Question G25 in 6th edition [ENDORSED]
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Question G25 in 6th edition
Question G25 in the sixth edition asks how many molecules of substance X are left after diluting a solution 90 times. What kind of equation can you set up to solve this?
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Re: Question G25 in 6th edition [ENDORSED]
I think the purpose of this question is based more on the concepts of dilution rather than applying the concepts. To find the concentration of a solution, I would still use mol/L where L is the total volume of the solution.
So here, since the question tells you that the solution is doubled each time, I believe the total volume at the end of 90 dilutions is 2^90 and the number of moles would be the initial amount (which can be calculated by multiplying the initial molarity of the solution by the 10.mL given to us in the problem.)
The final value would be so small(close to 0), that any amount of substance X would play an insignificant role in having any health benefits.
So here, since the question tells you that the solution is doubled each time, I believe the total volume at the end of 90 dilutions is 2^90 and the number of moles would be the initial amount (which can be calculated by multiplying the initial molarity of the solution by the 10.mL given to us in the problem.)
The final value would be so small(close to 0), that any amount of substance X would play an insignificant role in having any health benefits.
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Re: Question G25 in 6th edition
In order to find how many molecules remain after 90 doublings, you first have to calculate how many molecules are in the 10mL of solution:
.010 mL x (0.10 mol/L) x (6.022*10^23 molecules/1mol) = 6.0 x 10^20 molecules.
Then, you could set up an exponential formula:
6.0 x 10^22 molecules x (0.5^n)= # of molecules remaining, where n is the number of doublings.
Plug in 90 for n, and you get 0.000048468, which is almost nothing. Thus, the solution would be ineffective after diluting substance X to this extent.
.010 mL x (0.10 mol/L) x (6.022*10^23 molecules/1mol) = 6.0 x 10^20 molecules.
Then, you could set up an exponential formula:
6.0 x 10^22 molecules x (0.5^n)= # of molecules remaining, where n is the number of doublings.
Plug in 90 for n, and you get 0.000048468, which is almost nothing. Thus, the solution would be ineffective after diluting substance X to this extent.
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