# Numerical Questions – Beer Lambert Law: Questions and Answers

Dr. Manishika Jain- Join online Paper 1 intensive course. Includes tests and expected questions.

## Beer Lambert Law

Q1) In a spectrophotometric cell of 2.0 cm path length, the solution of a substance shows the absorbance value of 1.0. If the molar absorptivity of the compound is 2 x 10^{4} L mol^{-1} cm^{-1}, calculate the concentration of the substance in solution. What is the concentration of the substance in that solution?

(1) 2.5 x 10^{-5} mol L^{-1}

(2) 4.0 x 10-4 mol L-1

(3) 1.0 x 10^{4} mol L-1

(4) 5.0 x 10^{-4} mol L-1

Answer: (1) 2.5 x 10^{-5} mol L^{-1}

A = ε*C*L

Where,

Path Length (L) = 2.0 cm

Molar Absorptivity (ε) = 2 x 10^{4} L mol^{-1} cm^{-1}

Absorbance (A) = 1.0

1 = 2 x 10^{4} x C x 1

C =

C =

C = 2.5 x 10^{-5}

Q2). A solution of chemical ‘A’ having its 0.14 mol L−1 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 of ‘A’?

(1) 0.108 mol L−1

(2) 0.35 mol L−1

(3) 0.12 mol L−1

(4) 0.10 mol L−1

Answer: Option (3) 0.12 mol L−1

A = ε*C*L

Given Values,

Solution - 1

Absorbance = 0.42

Concentration= 0.14 mol

Solution - 2

Absorbance = 0.36

Concentration =?

For Solution 1

A= ε*C*L

0.42 = ε*0.14 *L ………….……………………………… (i)

For Solution 2

A = ε*C*L

0.36 = ε*C *L ………….……………………………… (i)

C = 0.12 mol L^{-1}

Q3) A student prepared a calibration curve by measuring of the absorbance of five standard solutions of a compound at 300nm. A cuvette with a path length of 5cm was used in the stratosphere the slope if the curve was 300 L/mol the molar absorptivity of the compound is

(1) 0.2 L mol^{-1}cm^{-1}

(2) 60 L mol^{-1}cm^{-1}

(3) 5.0 L mol^{-1}cm^{-1}

(4) 1500.0 L mol^{-1}cm^{-1}

Answer: Option (2) 60 L mol^{-1}cm^{-1}

Slope = 300 L/mol

Path Length = 5 cm

Molar Absorptivity =?

Slope = Molar Absorptivity x Path Length

300 = Molar Absorptivity x 5

Molar Absorptivity = 60 L mol-1cm-1

-Mayank