## Molarity and Dilution (G25)

Alyssa Pelak 1J
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### Molarity and Dilution (G25)

I can't seem to figure out this question...
Practitioners of the branch of alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is this claim plausible? To explore this question, suppose that you prepare a solution of a supposedly active substance, X, with a molarity of .10 mol L-1. 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.

Salman Azfar 1K
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### Re: Molarity and Dilution (G25)

This one can be tricky. You need to use the initial molarity and volume values they give you to calculate the number of moles, and from that, use Avogadro's number to get the number of atoms/molecules.
Once you have that, critical thinking comes in. Keep in mind that when you double the volume, you are cutting the number of molecules in each dose by a half each time. So you are repeatedly dividing it by 2. Acknowledging this fact means that you can set up a logarithmic expression to solve for how many cuts (doubling) it will take for each does to have only one molecule. Once you get your answer, you can tell how many molecules will be in each dose and whether or not some doses will even have a molecule or not. That will allow you to quickly comment on the effectiveness of the doses. If you need more specifics, let me know; I just didn't really want to go and write down all of the actual answers, so I thought I'd try to explain it step by step instead. Good luck!

Abigail Yap 2K
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### Re: Molarity and Dilution (G25)

My approach for this problem utilized the key concept that dilutions do not change the number of moles of solute in a solution. Assuming that the amount of X will remain the same after the 90 dilutions, I simply calculated for moles by multiplying the starting 0.10M molarity by the starting .010L volume and then converted this value to molecules by multiplying by Avogadro's number. This gave me 6.0 x 10^20 molecules X. Can someone please let me know if this is a valid way of solving? Thanks.

Daniel Vo 1B
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### Re: Molarity and Dilution (G25)

How I would go about that is to set up the equation between molarity, moles and volume. Starting with 10. mL (.010 L) and .10 mol L-1, we can set up

.10 M= .0010 moles / .010 L

So that'd be the initial equation. The independent variable we want to change is the volume, which we double 90 times (2^90), which will in turn change our molarity.

? M= .0010 moles / .010 L*2^90

Once you get the molarity, you can set up the equation one last time with your unknown being the moles, and the volume being .010 L once more. Whatever number you get will be very, very small.

I think.