165. LAWS OF PHOTOCHEMISTRY (11).

1.The molar absorptivity of a conjugated diene (molecular mass = 110 g/dm3) is 13.100.Calculate the concentration of this diene in a solvent, which is required to give absorbance of 1.6. The path length of light is 1.00 cm.

Solution –

Given – ε = 13.100 , A = 1.6 b= 1.00 cm , Molecular mass = 110g/mol.

According to Beer – Lambert’s Law –

A= εbc

Therefore , c = a/ ε b = 1.6/(13.100 *1) = 0.1221 mol/dm3.

2.Calculate the absorbance of a solution having a %T of 89 at 400nm.

Solution – We know,

A = 2 – log 10 (%T)
= 2 – log (89)
A = 0.051

3.The absorption co-efficient of a complex is 0.20 at light of wavelength of 450nm.What is the concentration when the transmission is 40% in a cuvette of 2 cm?

Solution –

Given – ε = 0.20 , %T = 40% = 0.4 , B = 2cm

We know, A= 2 – log (%T).

A = 2 – log (40) = 2 – 1.602 = 0.398

A = εbc

Therefore, c = A /ε b = 0.398/0.20 * 2 = 0.398 / 0.4 = 0.995.

Thus, the concentration of the complex is 0.995 mol/dm3

4.A solution of chemical “A” of 0.14mol/dm3 concentration has an absorbance of 0.42.Another solution of ‘A’ under the same conditions has an absorbance of 0.36.What is the concentration of this solution?

Solution-

By Beer- Lambert’s law , we know – A= ε cb.

ε is a constant , so it will change. The path length b is also not changing.

For solution 1 – A1= ε c1 b .Thus, ε b = A1 /c1 …..Equation(1)
For solution 2 – A2= ε c2 b .Thus, ε b = A2 /c2 …..Equation(2)

Equating equation 1 and 2 , we get ,


A1 /c1 = A2 /c2

Thus, c2 = (A2 * c1)/A1 = (0.36*0.14)/0.42 = 0.12 mol/dm3

The concentration of the second solution is 0.12 mol/dm3 .

We shall start discussing the second law of photochemistry next post onwards.Till then,

Be a perpetual student of life and keep learning….

Good day !

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