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Hi! There are actually multiple other ways to do this problem. So the first way is to do it manually, which is what we did in a TA office hour section I attended. Obviously you wouldn't do all 90 doubles manually, but after the first 2 or 3 you start to see that each subsequent double decreases by half and you can reasonably assume that by 90 doubles the value would be minuscule. What I did, however, was solve for the molarity and then convert it into a number of molecules, which I then multiplied by (1/2)^90. You get 4.9 x 10^-7 molecules per liter, which is tiny. Keep in mind that even with the manual's solution, at no point would the molarity actually be zero because the original moles of solute aren't changing--they're still in the diluted solution, they're just spread so far apart that they have virtually no effect on the solution anymore.
I think the method Ramya explained above does work and makes sense for G25, because it is not asking for when a specific number of molecules will be left in the dilutions. But it is important to understand the method with logarithms because if you look at G26, the problem might ask you to determine at what dilution will their be 1 molecule left. In such a scenario, you need to use logarithms to get the right answer.
Hi! I believe in Ramya's explanation of her way to solve the problem, she forgot to mention that she found the amount of moles in 10 mL of the solution, instead of 4.9 * 10^-7 molecules per Liter. I solved it in this fashion and got the same answer! Hope this helps.
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