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The absorbance A of a solution is defined as A=log 10 (Io/I) in which Io is the incident light intensity and I is the transmitted light intensity. The absorbance is related to the molar absorption coefficent e (in M-1 cm-1), concentration c (in M) and path length (in cm) by A=elc. The absorbtion coefficient of myoglobin at 580 nm is 15,000 M^-1 cm^-1.

a. What is the absorbance of a 1mg ml-1 solution across a 1-cm pathlength?
b. What percentage of the incident light is transmitted by this solution?

Respuesta :

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Answer:

a)- A= 0.84

b)- T= 14.4%

Explanation:

a) First, we need to convert the solution concentration (c) from mg/ml to mol/L by using the molecular weight of myoglobin, which is 17.8 kDa (17,800 g/mol):

c= 1 ml/ml = 1 g/L x 1 mol/17,800 g= 5.62 X 10⁻⁵ mol/L= 5.62 X 10⁻⁵ M

We have: e= 15,000 M⁻¹cm⁻¹, l= 1 cm, c= 5.62 X 10⁻⁵ M. Now, we can introduce the data in the expression for absorbance:

A= e x l x c = 15,000 M⁻¹cm⁻¹ X 1 cm x 5.62 X 10⁻⁵ M

A= 0.84

b) The percentage of the incident light which is transmitted by the solution is I/I₀.

We know that A= 0.84 and A= log₁₀(I₀/I), so:

0.84= log₁₀(I₀/I)

[tex]10^{0.84}[/tex]= I₀/I

6.97=I₀/I

We need the inverse (I/I₀):

I/I₀= 1/6.97= 0.144⇒ percentage: 0.144 x 100= 14.4 %

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