Question

| A solution containing 4.48 ppm KMnO4 exhibits

Answer 1

Answer:

Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹

Explanation:

First convert Concentration from ppm inM or mol/l

⇒ Molar mass of KMnO₄ = 158.03 g

⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l

⇒ Molarity = $\frac{4.48 X10^{-3} }{158.03X 1(lit)}$ = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T)     ( T = % transmittance)

                          = - log(0.859)

                          = 0.06

According to Lambert Beer's law

     

                 ε = $\frac{A}{C X l}$

      or,      ε = $\frac{0.06}{2.83 X 10^{-5}X1 cm }$

      or,      ε = 2120.14 cm⁻¹M⁻¹

Where

    ε = Molar absorptivity

    A = absorbance

    C = Molar concentration of KMnO₄ solution

     l = length