0.461 grams of a compound \"Z\" with a molecular weight of 154.12 g/mole was dis

Question

0.461 grams of a compound "Z" with a molecular weight of 154.12 g/mole was dissolved in water to make 50.0 mL solution. A 10.00 mL aliquot was withdrawn, placed in a 100.0 mL volumetric flask and diluted to the mark with water buffered at pH 5.40. 8.00 mL of this solution was transferred to a cuvette with a 1.00 cm path length, and the % Transmittance was measured at 470 nm to be 53.1 %. A second solution with an unknown concentration of "Z" has a transmittance of 29.3%. A. Calculate the molar absorptivity of "Z" at 470 nm. (don't forget units) B. Calculate the concentration of "Z" in the unknown solution.

Explanation / Answer

molar concentration of compound = moles/volume (L) of solution

                                                      = 0.461/154.12 x 0.05

                                                      = 0.06 M of "Z" in solution

10 ml of this solution was diluted to 100 ml

new molar concentration = 0.06 M x 10 ml/100 ml = 0.006 M of "Z" in solution

8 ml of this solution has transmittance = 53.1%

So,

absorbance = 2 - log(%T)

                    = 2 - log(53.1)

                    = 0.275

A. Molar absorptivity of "Z" at 470 nm = absorbance/concentration x path length

                                                            = 0.275/0.006 x 1

                                                            = 45.83 M-1.cm-1

B. Absorbance for the unknown = 2 - log(29.3)

                                                   = 0.533

So,

concentration of "Z" in the unknown solution = 0.533/45.83

                                                                       = 0.01163 M

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