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Inderpal Singh 2L
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Postby Inderpal Singh 2L » Fri Oct 02, 2020 2:40 pm

Can somehow please explain how the following is done:

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 molar concentration of
. 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.

I got the answer from a peer but the explanation did not sit well with me.

sophie esherick 3H
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Re: G25

Postby sophie esherick 3H » Fri Oct 02, 2020 2:52 pm

I approached this problem by using the M1V1 = M2V2 equation where M1 = initial Molarity, V1= initial volume, M2 = final molarity and V2= final volume. So (.10)(.01L) = (.02)(M2) therefore M2 = .05 M. This is half the original molarity therefore if you were to continue to do this 90 times, the final molarity would be very small so there would be little to no active X left in the 10mL sample of the solution. Without having a significant amount of active X in the 10mL solution, there are no health benefits.
I hope this made sense and helped you a bit!

John Pham 3L
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Re: G25

Postby John Pham 3L » Fri Oct 02, 2020 3:26 pm

If you were to double 10 mL 90 times, the solution would be equal to 10 * 2^90 mL. This exponent represents the doubling each time (i.e. 20, 40, 80... and so on) for 90 times. Using this value, I can recalculate the molarity. 10 * 2^90 mL is equal to approximately 1.24 * 10^25 L and there are 1.0 * 10^-3 moles in the initial 10 mL.

Recalculating the molarity after doubling the initial 10 mL 90 times would be equal to (1.0 * 10^-3 mol) / (1.24 * 10^25 L) which has a molarity of approximately 8.1 * 10^-23 mol/L. This value is a very small and insignificant value. If you were to take 10 mL of this solution with an insignificant molarity, the amount of X molecules remaining would be very few or none at all. As such, there would be no effect and no health benefit.

You wouldn't need to go through all the work to recalculate the molarity though. Like the other answer said, you are exponentially decreasing the molarity each time you double the solution which means after doubling 90 times, you would have basically no molecules left. Hope this made sense!

Nicoli Peiris 1B
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Joined: Wed Sep 30, 2020 10:02 pm

Re: G25

Postby Nicoli Peiris 1B » Mon Oct 05, 2020 2:39 pm

The key here is looking at what to plug in for volume final. Using M initial times V initial = M final times V final, we can plug in all given values using proper units to find M final. To do so we need to plug in a V final. We know that the initial volume is .010 L (once converted) and we know this amount is doubled 90 times. Rather than calculate .010 doubled manually 90 times, we can use an exponential function to represent final volume. This is done by taking the volume that will be doubled (0.010) and multiplying it by 2^90 since doubling is two and we are doing it 90 times. Using this we can find the Molarity which can lead us to moles which leads us to molecules. Hope this helps!

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