Intensity = 84,250 M-1 * [Li+] + 1.75
If the intensity of your unknown sample is 90, what is the concentration of Li+ in the analyzed sample?
90= 84,250 M-1 * [Li+] + 1.75
88.25 = 84,250 M-1 * [Li+]
1.05 E-3 M = [Li+]
If 5 mL of the original unknown sample was diluted to 200 mL prior to analysis, what is the concentration of Li+ in the original solution?
I need help with this question,
This is what I try,
1.05 E-3 M x 0.005 L = 5.25 E-6 mols
5.25E-6 mols x .200 L = 1.05 E-6 M =[Li]
2 answers
Too many "original" solutions in the problem. And I don't get the distinction between the unknown and the original sample and the analyzed sample. The dilution factor is 5/200 = 0.025.
I think what you did on first is right but the second on is wrong
you should use M1V=M2V1 formula
which M1(5mL)=(1.05*10^-3)(200mL)
M1=0.042
you should use M1V=M2V1 formula
which M1(5mL)=(1.05*10^-3)(200mL)
M1=0.042