Example:

The molar absorptivity of a substance is 2.0 × 10⁴ cm^-1 mol^-1 dm³. Calculate the transmittance through a cuvette of path length 5.0 cm containing 2.0 × 10^-6 mol dm ³ solutions of the substance.

Solution

As per the Lambert Beer's law,

Absorbance, A = ε c b

Given:

ε = 2.0 × 10⁴ cm^-1 mol^-1 dm³, c = 2.0 × 10^-6 mol dm ³ and b = 5.0 cm

Substituting the values we get,

A = 2.0 × 10⁴ × 2.0 10^-6 × 5.0 = 0.2

=> log l/T = 0.2        ∴ A = log 1/T

Taking antilog on both sides, we get l/T = 1.585

∴T = 0.63