Answer:
[tex] 1.35\times 10^{-4} M[/tex] was the concentration of the original solution.
Explanation:
[tex]M_1V_1=M_2V_2[/tex] (Dilution)
where,
[tex]M_1\text{ and }V_1[/tex] are molarity and volume of non diluted sample.
[tex]M_2\text{ and }V_2[/tex] are molarity and volume of diluted sample.
[tex]M_1=?[/tex]
[tex]V_1=100.0 \mu L=0.1 mL[/tex]
(1μL=0.001 mL)
[tex]M_2=6.41\times 10^{-6} M[/tex]
[tex]V_2=2.00 mL + 100.0 \mu L=2.00 mL+0.1 mL=2.1 mL[/tex]
Substituting all values :
[tex]M_1\times 0.1 mL=6.41\times 10^{-6} M\times 2.1 mL[/tex]
[tex]M_1=\frac{6.41\times 10^{-6} M\times 2.1 mL}{0.1 mL}=1.35\times 10^{-4} M[/tex]
[tex] 1.35\times 10^{-4} M[/tex] was the concentration of the original solution.