## G 25

Norman Dis4C
Posts: 101
Joined: Sat Sep 28, 2019 12:16 am

### 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 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? Comment on the possible health benefits of the solution."

The correct answer is no X remaining

Shouldn't the amount of solute, which is substance X remain the same?

Anna Wu 1H
Posts: 51
Joined: Sat Aug 17, 2019 12:16 am

### Re: G 25

The amount of solute remains the same, but the volume changes. Since the solution is being diluted so dramatically, the 10 mL sample taken at the end has close to zero solute in it.

Charisse Vu 1H
Posts: 101
Joined: Thu Jul 25, 2019 12:17 am

### Re: G 25

Since you are doubling the volume of the solution 90 times, the final volume would be much greater than the initial volume. The question asks how many molecules would be present in only 10. mL of the final solution. Since the volume has increased so dramatically, 10. mL of the final solution would contain a very, very minute amount of the molecule X.

Manav Govil 1B
Posts: 104
Joined: Sat Sep 07, 2019 12:19 am

### Re: G 25

The thing is that after 69 doublings, less than one molecule would remain in the 10 mL of the diluted solution. At that point, the solution does not have any health benefits whatsoever. When a solution has less than one molecule of a specific substance (especially if that substance is intended to aid the issue), the solution is ineffective, as the substance is too small to do anything. If you want exact numbers, I found that after 90 doublings, there are 4.9 * 10^-7 molecules in the solution - which is almost nothing.

Amir Bayat
Posts: 115
Joined: Sat Sep 07, 2019 12:16 am

### Re: G 25

I understand your guys' reasoning as to why there would be a very, very minute amount of X left in the substance. However, what is the fastest and quickest way to approach a question like this and show work?

I understand the conceptual and visualization aspect of it, but doubling until 69 times would take too long on a test. Have any of you guys found a way to approach this that would take less time?

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

### Re: G 25

I think once you see an absurd number that is above 50+, you can assume that you don't have to do all the math and just approach the question conceptually. If you use the concept of limits from algebra, if a number keeps getting smaller and smaller, you can safely make the assumption that it approaches 0 since it's so tiny.

Edmund Zhi 2B
Posts: 118
Joined: Sat Jul 20, 2019 12:16 am

### Re: G 25

In 20 mL of the 0.1M solution, there will be 0.1 = x/0.01, x = 0.001 mol of X (the solute). Every time you double this volume, there will be half as many moles in a 10 ml solution. The number of molecules in the final solution is therefore 0.001 mol * (1/2)^90 * avagadro's number, which is around 4.9 * 10^-7 molecules every 10ml, which is basically 0. There are no health benefits if the molecule is completely missing in the solution.

Amir Bayat
Posts: 115
Joined: Sat Sep 07, 2019 12:16 am

### Re: G 25

Update:

There is a much faster way to approach this and it is with using the formula below, once we have solved for the amount of molecules of X.

number of molecules of X solved for --> 6.0 * 10^20 molecules X + (1/2)^n = 1 molecule.

We then take the log of both sides and solve for n... getting 69 as the number of doubles to have one molecule of X left, and thus none left after 90 doubles.