Question

A student measures the absorbance of a sample of red dye #3 using a spectrophotometer. If the absorbance is measured as 0.468, what is the concentration of red dye #3 in the sample?

Enter your answer in units of micromolar to three significant figures.

n the lab spectrophotometer, d = 1.00 cm. For Red dye #3, ε = 7.96 x 104 M-1 cm-1.

Answer 1

Answer:

5.88 μM is the concentration of red dye in the sample.

Explanation:

Absorbance is defined s capacity of a substance to absorb a light of a specified wavelength.It is calculated by using Beer-Lambert Law:

$A=\epsilon lc$

A = absorbance

$\epsilon$ = Molar absorptivity

c = concentration

l =  Length of the solution through which light passes

Given, A = 0.468, l = d = 1.00 cm,c = ?

$\epsilon =7.96\times 10^4 M^{-1} cm^{-1}$

$A=\epsilon lc$

$0.468=7.96\times 10^4 M^{-1} cm^{-1}\times 1.00 cm\times c$

$c=5.8793\times 10^{-6} mol/L=5.8793 mol/(\mu L)\approx 5.88mol/(\mu L)$

($1 L= 1,000,000 \mu L$)

5.88 μM is the concentration of red dye in the sample.

Answer 2

Red dye #3 with a concentration of 5.88 μM has an absorbance of 0.468 in the lab spectrophotometer.

A student measures the absorbance of a sample of red dye #3 using a spectrophotometer and the value is A = 0.468. We can calculate the concentration of the ample using the Beer-Lambert law, which relates the attenuation of light to the properties of the material through which the light is travelling, through the following equation.

$A = \epsilon \times d \times c$

where,

• ε: molar absorptivity of the species (7.96 × 10⁴ M⁻¹ cm⁻¹)
• d: optical path length (1.00 cm)
• c: concentration

$c = \frac{A}{\epsilon \times d } = \frac{0.468}{7.96 \times 10^{4} M^{-1}cm^{-1} \times 1.00cm } \times \frac{10^{6} \mu M }{1M} = 5.88 \mu M$

Red dye #3 with a concentration of 5.88 μM has an absorbance of 0.468 in the lab spectrophotometer.

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