## HW G.25

ElizabethP1L
Posts: 59
Joined: Wed Nov 15, 2017 3:01 am

### HW G.25

Hey, everyone! I just had a general question about HW G.25. I understand the math behind it. However, the question asks if a very dilute solution of a substance X (presumably the solute) can still have an effect. I immediately thought yes because Dr. Lavelle mentioned that changing the volume of a solution doesn't change the number of solute molecules. However, the solutions manual shows that after 69 times of doubling the volume, there's no solute (Substance X) left. How can this be since dilution of a solution doesn't affect the solute? Thanks in advance!

Beverly Shih 1K
Posts: 34
Joined: Wed Nov 15, 2017 3:01 am
Been upvoted: 1 time

### Re: HW G.25

You're right that just diluting a solution doesn't change the total number of solute molecules in the solution, so after diluting the solution 90 times there is the same number of solute molecules you started with. But the question asks how many solute molecules would be present in 10. mL of the final solution, so you have to divide the volume of the final solution by some huge number to get 10. mL, and divide the original number of solute molecules by the same number which gives you less than one. This means if you divided the final solution into a lot of 10. mL beakers, there will probably be no molecules of the solute in most of the beakers, although some of beakers may each have one solute molecule that add up to the original number of solute molecules. But since the majority of the beakers would have no solute molecules, they wouldn't have any biological effect.