Question G.25

[404651793]

"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 molL1. 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 fi nal solution? Comment on the possible health benefits of the solution."

I am not sure how to double the volume of the solution 90 times without actually having to manually double it 90 times. What is the way to do it mathematically and in the solution it said there will be no more of the element x left to give any health benefits but how can that element be completely gone if it is a decay model and never actually reaches zero?

[Chem_Mod]

The final concentration would be 0.1 M * (0.5)^90

Every time you double the volume from 10 mL to 20 mL, the concentration simply goes to half. Then you do it 90 times with an exponent.

Then multiply the final molarity by 0.01 L * Avogadro's # to get the number of molecules

(there should not even be 1 molecule left)

[Erika Sosa-Cruz 1J]

Would light be considered a Quantum Mechanic because we could physically see the light. Or would it be a Classical Mechanic because of the energy that is giving of the light ?

[rita_h]

Upon dilution, moles are conserved.
Given: Co = 0.1 mol/L and Vo = 10 mL = 10 * 10^-3 L
n(initial) = n(final)
CoVo = CfVf --> Dilution factor (D) = Co/Cf = Vf/Vo
So, Dilution factor = 2^90 = Vf/ 10 (*2^90 because we are doubling the value 90 times)
Vf = 10 * 10 ^-3 * 2^90 = 1.238 * 10^25 L
If we rearrange CoVo=CfVf, we get Cf= CoVo / Vf = (0.1*10*10^-3) / 1.238 * 10^25 L
Cf = 8.078 * 10^-29 mol/L
and C = n/V
so n = C * V = 8.078* 10^-29 * 10 * 10^-3 = 8.078 * 10^-31
We know that N = n * Na = 8.078 * 10^-31 * Avogadro's cnst = 8.078 * 10^-31 * 6.022 * 10^23 which leads you to the number of molecules.
Based on that you can determine the effectiveness/possible health benefits of the doses.

[Chem_Mod]

Would light be considered a Quantum Mechanic because we could physically see the light. Or would it be a Classical Mechanic because of the energy that is giving of the light ?

This question is posted in the wrong section.
Quantum Mechanics posted in Molarity, Solutions, Dilutions ...

Questions posted in the wrong section are not answered.

Ask UAs (Monday onwards when their sessions start) how to post new questions in the correct section/topic.

[Anna Turk 1D]

Upon dilution, moles are conserved.
Given: Co = 0.1 mol/L and Vo = 10 mL = 10 * 10^-3 L
n(initial) = n(final)
CoVo = CfVf --> Dilution factor (D) = Co/Cf = Vf/Vo
So, Dilution factor = 2^90 = Vf/ 10 (*2^90 because we are doubling the value 90 times)
Vf = 10 * 10 ^-3 * 2^90 = 1.238 * 10^25 L
If we rearrange CoVo=CfVf, we get Cf= CoVo / Vf = (0.1*10*10^-3) / 1.238 * 10^25 L
Cf = 8.078 * 10^-29 mol/L
and C = n/V
so n = C * V = 8.078* 10^-29 * 10 * 10^-3 = 8.078 * 10^-31
We know that N = n * Na = 8.078 * 10^-31 * Avogadro's cnst = 8.078 * 10^-31 * 6.022 * 10^23 which leads you to the number of molecules.
Based on that you can determine the effectiveness/possible health benefits of the doses.

For this question, how did you know that you needed to use a dilution factor and where in the equation is the dilution factor used?

[rita_h]

Upon dilution, moles are conserved.
Given: Co = 0.1 mol/L and Vo = 10 mL = 10 * 10^-3 L
n(initial) = n(final)
CoVo = CfVf --> Dilution factor (D) = Co/Cf = Vf/Vo
So, Dilution factor = 2^90 = Vf/ 10 (*2^90 because we are doubling the value 90 times)
Vf = 10 * 10 ^-3 * 2^90 = 1.238 * 10^25 L
If we rearrange CoVo=CfVf, we get Cf= CoVo / Vf = (0.1*10*10^-3) / 1.238 * 10^25 L
Cf = 8.078 * 10^-29 mol/L
and C = n/V
so n = C * V = 8.078* 10^-29 * 10 * 10^-3 = 8.078 * 10^-31
We know that N = n * Na = 8.078 * 10^-31 * Avogadro's cnst = 8.078 * 10^-31 * 6.022 * 10^23 which leads you to the number of molecules.
Based on that you can determine the effectiveness/possible health benefits of the doses.

For this question, how did you know that you needed to use a dilution factor and where in the equation is the dilution factor used?

Because this question mentioned that we are doubling the volume of the solution 90 times (2^90), then the dilution factor is 2^90. A dilution factor is simply: Co/Cf = Vf/Vo. If you want to prove that it's true, you can try calculating Vf/Vo from the values I obtained => 1.238 * 10^25/10 * 10^-3 L = 1.238 * 10^27 = 2^90