Complete Question
49.9 ml of a 0.00292 m stock solution of a certain dye is diluted to 1.00 L. the diluted solution has an absorbance of 0.600. what is the molar absorptivity coefficient of the dye
Answer:
The value is [tex]\epsilon = 4118.1 \ M^{-1} cm^{-1}[/tex]
Explanation:
From the question we are told that
The volume of the stock solution is [tex]V_1 = 49.9 mL = 0.0499 \ L[/tex]
The concentration of the stock solution is [tex]C_1 = 0.00292 \ M[/tex]
The volume of the diluted solution is [tex]V_2 = 1.00 \ L[/tex]
The absorbance is [tex]A = 0.600[/tex]
Generally the from the titration equation we have that
[tex]C_1 * V_1 = C_2 * V_2[/tex]
=> [tex]0.00292 * 0.0499 = C_2 * 1[/tex]
=> [tex]C_2 = 0.0001457 \ M[/tex]
Generally from Beer's law we have that
[tex]A = \epsilon * l * C_2[/tex]
=> [tex]\epsilon = \frac{A}{ l * C_2 }[/tex]
Here l is the length who value is 1 cm because the unit of molar absorptivity coefficient of the dye is [tex]M^{-1} * cm^{-1}[/tex]
So
[tex]\epsilon = \frac{0.600}{ 1 * 0.0001457 }[/tex]
=> [tex]\epsilon = 4118.1 \ M^{-1} cm^{-1}[/tex]
