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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2017 · Paper I

Q6(b). Concentration from 80 % transmittance (Beer–Lambert law)

  • 10 marks
  • Answered
Question

Using a particular wavelength of light, a solution of a substance (molar mass 294 g mol⁻¹) transmits 80% of incident light. Determine the concentration of the solution in mol L⁻¹. Given that optical path length of the spectrophotometer tube is 1.0 cm and absorptivity of the substance is 2.0 L g⁻¹ cm⁻¹.

Answer in brief

A=-0.80=0.0969; c=A/ab=0.0485 g L⁻¹ =1.65×10⁻⁴ mol L⁻¹.

Key points

  • A=log(I₀)/(I)=abc, where a is the absorptivity (concentration in g L⁻¹) and ε=aM the molar absorptivity (concentration in mol L⁻¹).
  • Absorbance: A=-log T=-0.80=0.0969
  • c=A/ab=(0.0969)/((2.0)(1.0))=0.0485 g L⁻¹
  • c=0.0485g L⁻¹294g mol⁻¹ c=1.65×10⁻⁴mol L⁻¹

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Given: T=I/I₀=0.80; b=1.0 cm; a=2.0 L g⁻¹ cm⁻¹; M=294 g mol⁻¹.

  1. Beer–Lambert law: A=log(I₀)/(I)=abc, where a is the absorptivity (concentration in g L⁻¹) and ε=aM the molar absorptivity (concentration in mol L⁻¹).
  2. Absorbance: A=-log T=-0.80=0.0969
  3. Mass concentration: c=A/ab=(0.0969)/((2.0)(1.0))=0.0485 g L⁻¹
  4. Molar concentration: c=0.0485g L⁻¹294g mol⁻¹ c=1.65×10⁻⁴mol L⁻¹
  5. Same result with ε: ε=aM=588 L mol⁻¹ cm⁻¹; c=A/ε b=0.0969/588=1.65×10⁻⁴ mol L⁻¹.
  6. Note: transmittance is not linear in c; absorbance is. An absorbance of only 0.097 is at the low end for precise work (best between about 0.2 and 0.8).

Application. the same calculation underlies every colorimetric assay: iron as its 1,10-phenanthroline complex, phosphate as molybdenum blue, and proteins and DNA from their UV absorbance.

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Model answer to Q6(b), 2017 Paper I: solution page 26
Solution page 26 of 38
Read the whole 2017 solution2017 question paper

All parts of Question 6