G.25 Molarity of Solution after Diluting

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Layal Suboh 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

G.25 Molarity of Solution after Diluting

I'm having difficulty understanding question G. 25. I know that after doubling the volume the first time, the concentration will decrease to 0.5 mol.L^-1, but I don't know how to double the concentration 90 times without fully writing it out.

Practitioners of the branch of a alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is the claim possible? To explore this question, suppose that you prepare a solution of a supposedly active substance, X, with a molar concentration of 0.10 mol.L^-1. 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.

Thank you!

Srikar_Ramshetty 1K
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am

Re: G.25 Molarity of Solution after Diluting

A useful equation you could use is the exponential function and in this case - (10)(2)^90, which should give you a value of 1.23794 x 10^28 ml or 1.23794 x 10^25 L. Once you find the number of moles in the initial concentration you can use the Molarity equation to find your answer because the number of moles never changes and only volume does, as you have already figured out. M = n/1.23794 x 10^25 L

Hope this helps and makes sense.
- Srikar Ramshetty

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