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A solution, S1, of Yellow #5 has an absorbance far outside the linear region of Beer’s Law. In order to determine the concentration of this solution, a serial dilution is performed. To make solution S2, 20.00 mL of S1 is diluted to 100.00 mL. To make solution S3, 25.00 mL of S2 is diluted to 100.00 mL. The absorbance of S3 is measured to be 0.628 at 430 nm using a 1 cm cuvette. Use the true value of ε and calculate the concentration of solution S1.

Answer :

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Answer:

The concentration of solution S1 is 5.17 x 10⁻⁴ M.

Explanation:

Bee'rs Law: A = εcl

A: absorbance = 0.628

ε: molar absoptivity = 2.43 x 10⁴ M⁻¹cm⁻¹

c: molar concentration

l:pathlength = 1 cm

S3:

c = A/εl

c = 0.628/2.43 x 10⁴ M⁻¹cm⁻¹ x 1 cm

c = 2.58 x 10⁻⁵ M

2.58 x 10⁻⁵ mol Yellow #5 _______ 1000 mL

2.58 x 10⁻⁶ mol Yellow #5 _______  100 mL

S2:

2.58 x 10⁻⁶ mol Yellow #5 ______ 25.00 mL

                    x                      ______ 100.00 mL

                    x = 1.03 x 10⁻⁵ mol Yellow #5

S1:

1.03 x 10⁻⁵  mol Yellow #5 ______ 20.00 mL

                    x                      ______ 1000.00 mL

                    x = 5.17 x 10⁻⁴ mol Yellow #5

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