Problem G.25 in Fundamentals

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Alejandra Reyna 1E
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Joined: Wed Sep 21, 2016 2:58 pm

Problem G.25 in Fundamentals

Postby Alejandra Reyna 1E » Sun Jul 02, 2017 9:03 pm

The problem states: "To explore this question, suppose that you prepare a solution of a supposedly active substance, X, with a molarity 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?

I was hoping someone could clarify how I could approach this question. I became confused after the question asked for 90 doublings in all. Thank you in advance.

405060280
Posts: 24
Joined: Fri Feb 24, 2017 3:03 am

Re: Problem G.25 in Fundamentals

Postby 405060280 » Sun Jul 02, 2017 9:23 pm

One of the most important thing to remember when it comes to dilution is that the moles of the solute do not change before and after dilution. So in this particular problem, you just need to find out how many moles of substance X is in 10ml of the solution. Let me know if you have any questions.

Payton Schwesinger 1J
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Joined: Fri Sep 29, 2017 7:04 am

Re: Problem G.25 in Fundamentals

Postby Payton Schwesinger 1J » Tue Oct 03, 2017 9:59 pm

I was confused on this question as well. I understand that doubling it 90 times dramatically reduces the amount of solute per L of solution so that it almost becomes obsolete but I don't understand the solution manual's method of solving the problem. Could someone please help explain why they are trying to find out how many times the amount of molecules must be cut in half to equal one?

Scott Chin_1E
Posts: 55
Joined: Sat Jul 22, 2017 3:00 am

Re: Problem G.25 in Fundamentals

Postby Scott Chin_1E » Wed Oct 04, 2017 4:46 pm

I believe the reason why the solution manual is trying to prove the amount of doublings that would need to occur before there is only 1 atom of substance X is that after 70 doublings, the solution would essentially have 0 atoms of the substance, thus showing any more doublings past 70 doublings would not have any more active substance in the solution. So in other words, 70 marks the point at which the solution becomes useless, having lost all of its active substances.


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