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### Fundamentals #G25

Posted: **Thu Oct 03, 2019 4:38 pm**

by **Chris Tai 1B**

G.25 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 molar concentration of 0.10 M. 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? Comment on the possible health benefits of the solution.

How do I best set this up algebraically?

### Re: Fundamentals #G25

Posted: **Thu Oct 03, 2019 4:44 pm**

by **Joseph Saba**

If I understand correctly, I don't think there is a need to solve anything frankly. Dilution means that the amount of volume (most of the time water) is being increased without any change to the number of initial moles. The number of moles wouldn't change and therefore the X would still be the same as before.

### Re: Fundamentals #G25

Posted: **Thu Oct 03, 2019 4:54 pm**

by **Justin Seok 2A**

The number of molecules in the solution after all of the dilutions would be the same, however, the question asks for the number of molecules in 10 ml of soln, which would be now a small percentage of the total volume due to the ninety dilutions. So in order to solve for this, what I did was find the initial number of molecules in soln like so: .01 L * .1 mol/L * 6.0221*10^23 molecules/mol = 6.0*10^20 molecules of X. Then to find the number of molecules in 10 ml of soln after 90 dilutions, we multiply 6.0*10^20 by (1/2)^90, since every time we double the volume, the molarity of the solution is halved. In the end, it would 4.8467614e-7 molecules, basically meaning that there would be no molecules of X left since the amount is so much less than one.