Question

The molar absorption coefficient of a substance dissolved in benzene is measured to be 712 L mol-1 cm-1 at 403 nm. Calculate what fraction of the incident light intensity of that wavelength passes through 2.67 mm of a solution of concentration 3.17 mmol/L.

Answer 1

According to Beer-Lambert's law, the solution allows for the passage of 3/4 of the incoming light intensity at that wavelength.

Calculation:

According to Beer-Lambert's law,

A = ∈ * C * l

A = log(I₀/I)

T = I₀/I

log (I₀/I) = ∈ * C * l

Here,

A= absorbance of the solution

l= path length= 2.67 mm = 0.267 cm (∵ 1mm = 0.1 cm)

T= transmittance

C= concentration of the given solution = 3.17mmol/L = 3.17* 10⁻³ mol/L

I= transmitted light

I₀= incident light

∈= molar absorption coefficient = 712 L mol⁻¹cm⁻¹

A = ∈ * C * l= 712 * 3.17 * 0.267 *10⁻³ = 602.63* 10⁻³ =0.602

T = $10^{-A}$ = $10^{-0.602}$ = 0.25

so, the percentage of transmittance = 0.25 * 100 = 25 %

Therefore, the intensity reduction will be equal to the original radiation intensity less the transmittance.

Give the initial intensity = 100

So, the reduction = 100 - 25 = 75.

Hence the fraction of incident light intencity= 75/100 = 3/4

Therefore the answer is 3/4.

Learn more about Beer-Lambert's law here:

https://brainly.com/question/8831959

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