Please show all steps. (18 points) A 100 ppm solution of compound X gave an abso

Question

Please show all steps. (18 points) A 100 ppm solution of compound X gave an absorbance at 488 nm of 0.832 in a 1.00 cm cell, and a blank solution under the same conditions gave an absorbance of 0.007. When 1.248 g of unknown solid was dissolved in 1.000 L of solution, an absorbance reading of 0.709 was found 4. a. What are the true (adjusted) readings for the 100 ppm X and unknown X solutions? b. Using Beer's law, what is the absorption coefficient for compound X? Include the units. c. What is the concentration of X in the unknown solution? d. What is the mass of X in the unknown solution? e. What is the weight percent of X in the solid used to prepare the unknown solution?

Explanation / Answer

Ans. #a. True (adjusted) reading = Observed Abs of sample – Abs of blank

So,

            True (adjusted) reading for 100 ppm compound X = 0.832 – 0.007 = 0.825

            True (adjusted) reading for unknown X solution = 0.709 – 0.007 = 0.702

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

where,

                       A = Absorbance

                       e = molar absorptivity at specified wavelength (M-1cm-1)

                        L = path length (in cm)

                        C = Molar concentration of the solute

Since the molar mass of solute X is NOT mentioned, its ppm concentration can’t be converted to molarity. So, assuming a path length of 1.0 cm, calculate absorption coefficient, e in terms of given concentration.

Putting the values in above equation –

            e = A / CL = 0.825 / (100 ppm x 1.0 cm)

            Hence, e = 0.00825 ppm cm-1

#c. Putting the value of unknown’s abs and e in equation 1-

            C = A / e L = 0.702 / (0.00825 ppm cm-1 x 1.0 cm) = 85.091 ppm

#d. Mass of X in unknown soln. = [X] in unknown soln. x Vol. of soln. in liters

                                                = 85.091 ppm x 1.0 L

                                                = 85.091 mg L-1 x 1.0 L

                                                = 85.091 mg

                                                = 0.0851 g

#e. % X (wt/wt) in sample = (Mass of X / Mass of sample) x 100

                                    = (0.0851 g / 1.248 g) x 100

                                    = 6.82 %

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