Question

Calculate the expected absorbance of solution of a 5.50x10-6 M solution of Blue # measured at 629 nm in a 1.50cm cell. The molar absorptivity for Blue #1 is 1.30×105 M-1cm-1 at 629 nm 4. A 2.50x10-5 M solution of a specific dye has an absorbance value of 1.300 when measured at 525 nm in a cell with a path length of 2.00 cm. Calculate the molar absorptivity of this dye at this wavelength 5.

Answer 1

Ans. #4. Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                                                A = Absorbance

                                                e = molar extinction coefficient (M^-1cm^-1)

                                                L = path length (in cm)

                                                C = Molar concentration

Putting the values in equation 1-

            A = (1.30 x 10⁵ M^-1cm^-1) (5.50 x 10^-6 M^-1) x 1.50 cm

Hence, A = 1.07250

#5. Putting the values in equation 1-

            1.300 = e x (2.50 x 10^-6 M^-1) x 2.0 cm

Or, e = 1.300 / (2.50 x 10^-6 M^-1 cm^-1)

Hence, e = 2.600 x 10⁵ M^-1cm^-1

#1. The frequency of an electromagnetic radiation or photon is given by the equation-

                        c = v l                      - equation 2

            Where, c = speed of light = 299792458 m/s

                        v = frequency            ; l = wavelength

Putting the value of wavelength in equation 2-

                        v = c / l

                        Or, v = (299792458 m/s) / (4.20 x 10^-7 m)                        ; [1 m = 10⁹ nm]

                        Hence, v = 7.1379 x 10¹⁴ s^-1

# Energy of the photon is given by-

                        E = hv            - equation 1

; where, h = Plank’s constant = 6.626 x 10^-34 Js   ; v = frequency of photon   

Putting the value of v in above equation-

            E = (6.626 x 10^-34 Js) x (7.1379 x 10^-14 s^-1)

            Hence, E = 7.7296 x 10^-19 J

#2. % Transmittance = 100 % - % Absorbance = 100 % - 68.5 % = 31.5 %

Now,

            Absorbance = 2 – log (%T) = 2 – log 31.5 = 0.5017                 

#3. Molar mass of the dye, C₃₇H₃₄N₂Na₂O₉S₃ = 792.862576 g/ mol

Mass of dye taken = 55.0 mg = 0.055 g

Volume of stock solution made upto = 250.0 mL = 0.250 L

Moles of dye taken = Mass / molar mass

                                    = 6.9369 x 10^-5 mol

Molarity of stock solution = Moles of dye / Volume of solution in liters

                                    = 6.9369 x 10^-5 mol / 0.250 L

                                    = 2.7748 x 10^-4 mol/ L                                

# Working solution: Using -

            C1V1 (stock solution) = C2V2

            Or, 2.7748 x 10^-4 mol L^-1 x 10.0 mL = C2 x 500.0 mL

            Or, C2 = (2.7748 x 10^-4 mol L^-1 x 10.0 mL) / 500.0 mL

            Hence, C2 = 5.5495 x 10^-6 mol/L

Hence, [dye] in working solution = 5.5495 x 10^-6 mol/L