Problem G.25
Posted: Fri Oct 05, 2018 4:51 pm
So this problem was really easy at the beginning, but it quickly turned hard at the end. The question is
" G.25 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 substance, X, with a molarity of 0.10 mol/L. 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. "
What I got was:
molarity of x= 0.10 mol/L
10 mL * .10 mol/1000 ml * (6.022 * 10^23)/ 1 mol = 6.022 * 10^20 molecules in first 10 mL
I Initially thought this was my final answer until I double checked with the solution manual and saw that it wasn't.
What I Need help in is explaining the meaning or understanding of how you know what to use to find the final solution?
Second Part of problem:
6.0 * 10^20 * (1/2) ^n = 1 molecule
log(6.0 * 10^20) * nlog(1/2) = log(1)
20.8 + n*(-.30) = 0
20.8= .30n
20.8/.30= n
n= 69
" G.25 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 substance, X, with a molarity of 0.10 mol/L. 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. "
What I got was:
molarity of x= 0.10 mol/L
10 mL * .10 mol/1000 ml * (6.022 * 10^23)/ 1 mol = 6.022 * 10^20 molecules in first 10 mL
I Initially thought this was my final answer until I double checked with the solution manual and saw that it wasn't.
What I Need help in is explaining the meaning or understanding of how you know what to use to find the final solution?
Second Part of problem:
6.0 * 10^20 * (1/2) ^n = 1 molecule
log(6.0 * 10^20) * nlog(1/2) = log(1)
20.8 + n*(-.30) = 0
20.8= .30n
20.8/.30= n
n= 69