lab report redox titration theory
}\, \text{mol} \] Since the ratio is 1:1, \[ \text{Moles of analyte} = 4.0 \times 10^{-4}\, \text{mol} \] The molarity of analyte: \[ M_2 = \frac{\text{moles of analyte}}{\text{volume of analyte}} = \frac{4.0 \times 10^{-4}}{0.02