G.25 -- Homeopathy Question

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Ben Rolnik 1D
Posts: 33
Joined: Fri Jun 23, 2017 11:39 am

G.25 -- Homeopathy Question

Postby Ben Rolnik 1D » Thu Jun 29, 2017 12:44 pm

Problem G.25 says we start with .10M of Substance X.... so .10 = (n / v). I got that part!

The next part of the problem seems to be saying something like: " .10M * (10 mL) = [2^90 * ( 10 mL) ] * [final molarity Molarity]

At this point I get kind of lost... especially because the answer in the solutions manual does this problem in a very different way!

What am I misreading about this problem? And how can I conceptualize how to get to the correct answer?


Kyle Sheu 1C
Posts: 87
Joined: Fri Jun 23, 2017 11:39 am

Re: G.25 -- Homeopathy Question

Postby Kyle Sheu 1C » Thu Jun 29, 2017 5:35 pm

Your understanding of the problem seems correct, and the way the solution manual approaches it errs of the side of dubiousness (more on this later).
Here's what I did:
1) 0.1M and a 0.01L sample
2) gives us 0.001 moles of X.
3) We take our 10 mL and...
double it 1x: V=10*2=20mL
double it 2x: V=20*2=40mL
double it 3x: V=40*2=80mL
double it "n" times: V=10*2n
Therefore, doubling it 90 times: V=10*290=1.23794*1028 mL

4) Law of Conservation of Matter tells us that the 0.001 moles of X in the original sample are still in this highly dilute solution.

5) The question asks how many molecules are in a 10 mL sample of this dilute solution, so we will set up a proportion to determine the number of moles in the dilute solution:
0.001mol/1.23794*1028mL = y mol/10mL (where y is an unknown)
y=8.0779*10-31 moles

6) There are 8.0779*10-31 moles in 10mL of the highly diluted solution, so we now convert to molecules:
8.0779*10-31 moles * (6.023*1023molecules/1 mole)=4.865*10-7 molecules (!!!)

What this means is that, mathematically speaking, in 10mL of this dilute solution, there is only a very small fraction of molecule X! Realistically, there are 0 molecules of X in 10mL, thus rendering the dilute solution useful only for ion-free drinking water.
What the solution manual does is check from the outset if, after 90 dilutions, there will be, on average, any molecules in 10mL of the dilute solution. Using their logarithmic equation, they determine that the maximum number of dilutions that can be done such that there is at least, on average, one molecule of X in 10mL of dilute solution (this number is deemed to be 69).

Let me know if there is anything I can clarify! This was a great problem!

Ben Rolnik 1D
Posts: 33
Joined: Fri Jun 23, 2017 11:39 am

Re: G.25 -- Homeopathy Question

Postby Ben Rolnik 1D » Sun Jul 02, 2017 3:18 pm


What an awesome response -- thank you so much for this!!! I totally understand it now. Thank you :)

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