## G25 homework problem

Kaylin Krahn 1I
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### G25 homework problem

The question: Practitioners of the branch of alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is the claim plausible? To explore this question, suppose that you prepare a solution of a supposedly active substance, X, with a molarity of 0.10 mol/L. Then you dilute 10. mL of that solution by doubling the volume, doubling it again, and so on, for 90 doublings in all. How many molecules of X will be present in 10. mL of the final solution?

So I multiplied (.01 L) x (.10 mol) x (6.02214x10^23 molecules)= 6.0x10^20 molecules

How do I go from this point?

Nathan Tu 2C
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Joined: Fri Sep 29, 2017 7:07 am

### Re: G25 homework problem

Well since you found out how many molecules you have, it is important to note that is your initial amount of molecules. Since initial amount should equal final amount, because number of moles/ molecules should not change when diluting a solution, the initial amount is the final amount. Thus your 6.02 x 10^20, I think that what it was, should be the final amount of molecules. Hope this helps. By the way, can someone help with the second part? I know the answer is no, because by constantly diluting the solution, you decrease the molarity which naturally means that one must drink more solution to receive a sufficient amount of moles the substance. But the answer book has something completely different and I don't understand the way they explain it. Can someone please help?

Ramya Natarajan 1D
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### Re: G25 homework problem

Hey there! You know how many molecules you have, so that's a great start, and you should multiply that by (1/2)^90 because you doubled the number you had all the way to 90 dilutions:

6.022e20 (1/2)^90 = 0.000000048 molecules, or essentially, so insignificant that there can be said to be no molecules at all. Therefore, there are no health benefits to X since there are no active molecules left in the substance.

Ivy Lu 1C
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### Re: G25 homework problem

I don't still don't really get how to do this problem. Can someone explain how to do it more?

Posts: 88
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### Re: G25 homework problem

I also don't understand the problem very well. Do you think it'll be mentioned in the test on Friday?

Ethan Mondell 1A
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### Re: G25 homework problem

Are there additional ways to solve this problem other than the solutions listed in the manual and this forum?

Gurkriti Ahluwalia 1K
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Joined: Fri Sep 29, 2017 7:07 am

### Re: G25 homework problem

I actually approached this problem a bit more practically in terms of limits. I tested the doublings in the formula MinitialVinitial=MfinalVfinal sort of like this:

(.1M)(.010L)=(xM)(.020L) where the final concentration is .05 M. This is just the FIRST doubling. as expected, the concentration became half by the first doubling.

so you double 20 which is 40, and double 40 which is 80, and double 80 which is 160, and so on and so forth.

you can test this again this time by using .160 instead of .020 and the concentration will be .00625. test this doubling process again by using 1280 (the seventh tier double) instead of .020 and you will get .0007812 M.

notice how the M is increasing zeroes after the decimal. In other words, the concentration is getting closer to zero. So I can assume that by the time the 90th doubling has come, there is insignificantly very little left, which will overall have no effect. No concentration? No effect.

Connor Kelligrew 2D
Posts: 39
Joined: Fri Sep 29, 2017 7:07 am

### Re: G25 homework problem

Ivy Lu 1C wrote:I don't still don't really get how to do this problem. Can someone explain how to do it more?

Well I started by reading the problem and seeing I have the molarity and the volume of the solution. So I can use the equation: $M= \frac{n}{V}$
I want to find the moles of X in the solution, so I will switch some things around: $n=MV$
Then plug in values: $n=(0.10mol.L^{-1})(0.010L)=0.001mol$
So that is the number of moles of X in the initial solution, which you can find the molecules of.

I found the molecules of X present in 10. mL of the final solution differently than the solution manual. Doubling the volume of solution each time can be represented as 2 to the power of however many times the volume is doubled times the initial volume of solution. The moles of solute in a solution never changes no matter how much extra solvent is added, so we can keep our result from above while calculating the new molarity of the 90 times doubled solution: $M_{f}=\frac{0.001mol}{2^{90}(0.01L)}$
Which gives a final molarity of 8.078 x 10^-29. Multiply this by 10. mL as done earlier to get the moles in 10. mL of solution, then multiply this by Avogadro's Number to get the number of molecules, which would be 4.86 x 10^-7.
This result is extremely small, so it's safe to say the active ingredient, X, will have no effect when 10. mL of the diluted solution is given to a patient, and so it won't provide health benefits. This is basically how someone stated it above, just extended to include some details. Hope it helps!