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]
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
To find the concentration of Li+ in the original solution, you need to take into account the dilution factor.
First, let's calculate the number of moles of Li+ in the 5 mL of the unknown sample:
1.05 E-3 M (concentration of Li+) x 0.005 L (volume of the sample) = 5.25 E-6 moles Li+
Next, let's calculate the concentration of Li+ in the diluted solution:
5.25 E-6 moles Li+ / 0.200 L (dilution volume) = 2.625 E-5 M
Therefore, the concentration of Li+ in the original solution is 2.625 E-5 M.