Problem G.25

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Hannah Morales 1D
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am

Problem G.25

Postby Hannah Morales 1D » Fri Oct 05, 2018 4:51 pm

So this problem was really easy at the beginning, but it quickly turned hard at the end. The question is

" 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. "

What I got was:

molarity of x= 0.10 mol/L

10 mL * .10 mol/1000 ml * (6.022 * 10^23)/ 1 mol = 6.022 * 10^20 molecules in first 10 mL

I Initially thought this was my final answer until I double checked with the solution manual and saw that it wasn't.

What I Need help in is explaining the meaning or understanding of how you know what to use to find the final solution?

Second Part of problem:

6.0 * 10^20 * (1/2) ^n = 1 molecule

log(6.0 * 10^20) * nlog(1/2) = log(1)
20.8 + n*(-.30) = 0
20.8= .30n
20.8/.30= n
n= 69

Michael Torres 4I
Posts: 92
Joined: Thu May 10, 2018 3:00 am
Been upvoted: 1 time

Re: Problem G.25

Postby Michael Torres 4I » Fri Oct 05, 2018 11:58 pm

Hannah Morales 1D wrote:So this problem was really easy at the beginning, but it quickly turned hard at the end. The question is

" 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. "

What I got was:

molarity of x= 0.10 mol/L

10 mL * .10 mol/1000 ml * (6.022 * 10^23)/ 1 mol = 6.022 * 10^20 molecules in first 10 mL

I Initially thought this was my final answer until I double checked with the solution manual and saw that it wasn't.

What I Need help in is explaining the meaning or understanding of how you know what to use to find the final solution?

Second Part of problem:

6.0 * 10^20 * (1/2) ^n = 1 molecule

log(6.0 * 10^20) * nlog(1/2) = log(1)
20.8 + n*(-.30) = 0
20.8= .30n
20.8/.30= n
n= 69


The following method is how I solved this problem.

Molarity = (moles)/(volume in Liters)

Therefore, 0.10M = 0.10 moles/1L.

Now, take 10mL of the solution and place it into its own container. The molarity remains the same, but the number of moles changes to compensate for the change in volume.

0.10M = 0.0010 moles/0.01L

Now, double the volume ninety times. Pay attention to how this changes the molarity.

(0.0010 moles)/(0.01L * 2^90) = 8.08 * 10^-29M

Now, you set up the equation to figure out how many moles of the substance remain in the solution.

8.08 * 10^-29M = ? moles/0.01L

The resulting number of moles is 8.08 * 10^-31.

The resulting number of moles is very small. It demonstrates there will be no molecules of the substance left in the solution, and as such the health benefits of the solution are negligible.


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