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