Ergosterol, a precursor of vitamin D, has δmax = 282 nm and molar absorptivity ε = 11,900 M-1cm-1. What is the concentration of ergosterol in a solution whose absorbance A = 0.93 with a sample pathlength l = 1.00 cm. M. What is the absorbance of a solution having a concentration of ergosterol of 4.1×10-5 M with a sample pathlength l = 1.00 cm?

Respuesta :

Answer:

1) [tex]7.8\times 10^{-5} M[/tex] is the concentration of ergosterol in a solution.

2) The absorbance of the solution is 0.49.

Explanation:

Using Beer-Lambert's law :

Formula used :

[tex]A=\epsilon \times C\times l[/tex]

where,

A = absorbance of solution

C = concentration of solution = [tex]2.00\times 10^{-3}M[/tex]

l = path length = 1.00 cm

[tex]\epsilon[/tex] = molar absorptivity coefficient

1)

We have:

Absorbance of the solution ,A = 0.93

Concentration of ergosterol ,C = ?

Path length ,l = 1.00 cm

[tex]\epsilon = 11,900 M^{-1} cm^{-1}[/tex]

[tex]A=\epsilon \times C\times l[/tex]

[tex]C=\frac{A}{\epsilon\times l}[/tex]

[tex]=\frac{0.93}{11,900 M^{-1}cm^{-1}\times 1.00 cm}=7.8\times 10^{-5} M[/tex]

[tex]7.8\times 10^{-5} M[/tex] is the concentration of ergosterol in a solution.

2)

We have:

Absorbance of the solution ,A = ?

Concentration of ergosterol ,C = [tex]4.1\times 10^{-5} M[/tex]

Path length ,l = 1.00 cm

[tex]\epsilon = 11,900 M^{-1} cm^{-1}[/tex]

[tex]A=\epsilon \times C\times l[/tex]

[tex]=11,900 M^{-1}cm^{-1}\times 4.1\times 10^{-5}M\times 1.00 cm= 0.49[/tex]

The absorbance of the solution is 0.49.