Question

C. Problems 1Explain the difference between transmittance, absorbance, and molar absorptivity Which one is proportional to concentration? 2. The absorbance of a 2.3lx 10 M solution of a compound is 0.822 at a wavelength of 266 nm in a 1.00 cm cell. Calculate the molar absorptivity at 266 nm 3. A compound with a molecular weight of 292.16 was dissolved in a 5-ml volumetric flask. A 1.00-mLaliquat was withdrawn, placed in a 10-mL volumetric flask, and diluted to the mark. The absorbance measured at 340 nm was 0.427 in a 1.000-cm cuvette. The molar absorptivity for this compound at 340 nm is a340-6130 M-'em a) Calculate the concentration of the compound in the cuvette b) What was the concentration of compound in the 5-mL flask? c) How many milligrams of compound were used to make the 5-mL solution?

Answer 1

C.

1.

Transmittance : It is the fraction of incident light that is not absorbed by the sample .

Transmittance (T) =

Io 0

I is intensity of light after passing through the sample.

I_o is intensity of incident light.

Absorbance (A) : It is the capacity of a substance to absorb light.

It is expressed as ; A = log (1/T).

Molar absortivity ($\epsilon$)  : Molar absortivity measures how strongly a substance absorbs light.

By Lambert Beer's law

A = $\epsilon$ * c * l

Where, c is concentration

l is cell lemgth.

So, absorbance is proportional to concentration.

2.

By Lambert Beer 's law

A = $\epsilon$ *c*l

Given, c = 2.31*10^-5 M, l = 1.00 cm

A = 0.822

Then, molar absortivity ($\epsilon$) = A/c*l = (0.822/2.31*10^-5*1)

= 3.55*10⁴ M^-1 cm^-1.

3.

Again, A = $\epsilon$ *c*l

Given, $\epsilon$ = 6130 M^-1 cm^-1

l = 1.000 cm

A = 0.427

Then , concentration (c) =( A/$\epsilon$*l)

= (0.427/6130*1.000)

= 6.96*10^-5 M.

b)

Final Volume (V₂) = 10.00 ml

Final concentration (M₂) = 6.96*10^-5 M.

Initial volume (V₁) = 1.00 mL

Let, Initial concentration = M₁

Then, M₁V₁ = M₂V₂

Or, M₁ = (M₂V₂/M₁) = (6.96*10^-5*10.00/1.00)

= 6.96*10^-4 M.

So, concentration of the compound in 5 ml flask

= 6.96*10^-4 M.

c)

Molarity = milimoles of solute/ volume (mL)

Or, milimoles of solute = Molarity * volume(L)

= 6.96*10^-4 * 5 = 3.48*10^-3 .

Now, mass of the compound (solute) in miligram

= Milimoles of the compound * molar mass

= 3.48*10^-3 * 292.16 = 1.016 mg.

So, 1.016 mg of the compound was used.