G. 25

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Brittany Tran 3I
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G. 25

Postby Brittany Tran 3I » Thu Oct 03, 2019 9:45 pm

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.

Can someone explain how to do this problem? the solution guide used logs but I don't understand why.

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Re: G. 25

Postby Chem_Mod » Thu Oct 03, 2019 9:58 pm

Logarithms are used to solve for unknowns that are the exponent of something (in this case 1/2 to the nth power). Logarithms allow us to easily solve for n. Refreshing on your logarithm rules would be helpful!

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Re: G. 25

Postby KnarGeghamyan1B » Thu Oct 03, 2019 11:19 pm

For this question, if you think about it logically instead of trying to solve equations, there would be pretty much no molecules of x left in the solution. You start off with a certain amount of a substance that is dissolved, which means the molecules are evenly distributed throughout the liquid, unless we're told differently. So when we dilute the solution 90 times over, there is still the same amount of moles of the substance, but it is spread out throughout a much larger volume. This relates to the initial moles=final moles phenomenon we learned in class. Therefore, the solution is so diluted that when you take a 10ml sample from the massive new solution, there will be only a small fraction of the moles you started with.

Hussain Chharawalla 1G
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Re: G. 25

Postby Hussain Chharawalla 1G » Thu Oct 03, 2019 11:28 pm

To add on, think of it intuitively as salt and water. If you have a 10 grams of salt in 10 mL of water and then double the water amount 90 times you would eventually have 10 grams of salt now dispersed in a huge body of water (10*(2)^90) mL. So if you were to take a small sample of this new diluted solution the amount of salt you would find in there would not be the original amount but rather a small fraction of it. Try to apply that thinking to this problem.

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