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

Answer:

1.10  × 10³ L/mol.cm

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:

$\frac{3.92 mg}{100mL}*\frac{1g}{10^3mg}*\frac{10^3mL}{1L}*\frac{1molA}{335gA}$

= 1.170 × 10⁻⁴ mol/L

Now, from equation (1)

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

ε = $\frac{0.1931}{1.50*1.170*10^{-4}}$

ε = 1100.02849 L/mol.cm

ε =  1.10  × 10³ L/mol.cm

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