Question

A solution containing 3.92 mg/100 mL of A (335 g/mol) has a transmittance of 64.1% in a 1.50-cm cell at 425 nm. Calculate the molar absorptivity of A at this wavelength.

Answer 1

1.10 × 10³ L/mol.cm

Step-by-step explanation:

Formula for Beer's law showing the relationship between concentration and absorbance is given as:

A = εbc -------- Equation (1)

where:

A = absorbance

ε = molar absorptivity

b = path length

c = concentration of analyte in moles/ Litre

Given that:

Trasnsmittance = 64.1%

we can find the value for Absorbance A = - log T

Absorbance (A) = -log (64.1%)

Absorbance (A) = -log (0.641)

Absorbance (A) = 0.1931

the path length of the cell = 1.50-cm

concentration of Solution A = 3.92 mg/100 mL

Molar Mass of A = 335 g/mol

Let convert our data given from mg/mL to mol/L, then substituting them into equation (1) to determine the value of our molar absorptivity.

So; we have:

`(3.92 mg)/(100mL)*(1g)/(10^3mg)*(10^3mL)/(1L)*(1molA)/(335gA)`

= 1.170 × 10⁻⁴ mol/L

Now, from equation (1)

0.1931 = ε(1.50cm)(1.170 × 10⁻⁴ mol/L)

ε =
`(0.1931)/(1.50*1.170*10^(-4))`

ε = 1100.02849 L/mol.cm

ε = 1.10 × 10³ L/mol.cm

∴ The molar absorptivity of A = 1.10 × 10³ L/mol.cm