Fundamental G25

[Jessy Ji 2J]

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 sub- stance, X, with a molar concentration of 0.10 mol?L21. 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.

For this question, I know that there will be basically no molecules of X left because it is so diluted. However, is it acceptable to just write that or should I calculate and write an actual number showing how many molecules of X will be present in 10 mL of the final solution? And if I have to calculate a number, how would I do that?

[Stephen_Kim_1D]

I'm not sure if you have to show your calculations for this type of problem but if I had to I would probably show my work as simply as I can:

In 10 mL of a 0.10 mol/L solution, there are 0.001 moles of X.
The molarity after 90 doublings would be (0.001 moles)/(0.01*2^90). A small number divided by an extremely large number is roughly 0.
10 mL of a 0 M solution realistically contains 0 moles of X, so there are hardly any molecules of X left.

[Jessy Ji 2J]

Thank you for answering my question! but could you explain a little more about why is it (0.001 moles)/(0.01*2^90)? I didn't really know where to get 0.01*2^90.

[Alison King 3L]

You are doubling the volume 90 times. If you doubled the volume twice it would be .01*2*2. Doubling 4 times is .01*2*2*2*2 or .01*2^4. Hence doubling 90 times is .01*2^90

[Chem_Mod]

Hello all!

Stephen and Alison's explanations are correct! To determine the number of moles in the 10 mL solution, first convert the 10 mL volume to liters. Then, use the molarity (0.1 mol/L) to determine the number of mols in the 10 mL solution. To determine the molarity after doubling the solution, divide your number of mols by your new volume (20 mL). For "n" doublings of the volume, the original volume 0.01 L is multiplied by 2 to the power "n". When the volume is doubled 90 times, the mols present in the solution becomes essentially 0. See attachment below.

-Gabby (TA for 1E, 1K, and 1L)