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UPSC Combined Geo-Scientist

Chemist

  • 3 descriptive papers
  • 200 marks each
  • 3 h per paper

Mains Paper III covers analytical and organic chemistry.

UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper III

Q1(h). MnO₂ in pyrolusite by back titration

  • 5 marks
  • Topic: Redox titrations
  • Answered
Question

A solution comprising 1.0 g pyrolusite is treated with 100 mL of 0.1 N oxalic acid and H₂SO₄. The excess oxalic acid requires 27 mL of 0.05 N KMnO₄ for complete titration. Find the amount of MnO₂ (in wt%) in the sample. [Given: molar mass of MnO₂ = 87 g mol⁻¹]

Answer in brief

MnO₂ =37.6 %.

Key points

  • Oxalic acid taken: 100×0.1=10.00 meq.
  • Left over (equal to KMnO₄ used): 27×0.05=1.35 meq.
  • Used by MnO₂: 10.00-1.35=8.65 meq.
  • Mass of MnO₂ =8.65×10⁻³×43.5=0.3763 g % MnO₂=(0.3763)/(1.0)×100=37.6 %

The full model answer

Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.

MnO₂ + H₂C₂O₄ + 2H⁺ → Mn²⁺ + 2CO₂ + 2H₂O (equivalent mass of MnO₂ =87/2=43.5)

  1. Oxalic acid taken: 100×0.1=10.00 meq.
  2. Left over (equal to KMnO₄ used): 27×0.05=1.35 meq.
  3. Used by MnO₂: 10.00-1.35=8.65 meq.
  4. Mass of MnO₂ =8.65×10⁻³×43.5=0.3763 g % MnO₂=(0.3763)/(1.0)×100=37.6 %
Model answer to Q1(h), 2026 Mains Paper III: solution page 7
Solution page 7 of 30
Read the whole 2026 solution2026 question paper

Coversback titrationMnO₂ in pyrolusite by back titration

All parts of Question 1