UPSC Combined Geo-Scientist, Mains · 2024 · Mains Paper III
Q1(c). Transmittance from Beer’s law
Question
A 4.00 × 10⁻⁴ M solution of aminobenzene in water has absorbance, A = 0.504 at 280 nm in a 1.00 cm cell. What will be the transmittance of a 1.50 × 10⁻³ M solution of aminobenzene in water measured at the same wavelength but in a 0.5 cm cell?
Answer in brief
ε = 1260 L mol⁻¹ cm⁻¹; new absorbance A = 0.945, so T = 10⁻⁰·⁹⁴⁵ = 0.114 (11.4 %).
Key points
- Beer–Lambert law: A = εbc = −log T; ε is the same for both solutions (same substance, solvent and wavelength).
- ε = A₁/(b₁c₁) = 0.504/(1.00 × 4.00 × 10⁻⁴) = 1260 L mol⁻¹ cm⁻¹.
- A₂ = εb₂c₂ = 1260 × 0.5 × 1.50 × 10⁻³ = 0.945.
- T = 10⁻ᴬ² = 10⁻⁰·⁹⁴⁵ = 0.114 (%T = 11.4); the first solution has T = 0.313 (31.3 %).
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversBeer–Lambert lawtransmittanceabsorbancemolar absorptivityanilineUV-Vis
All parts of Question 1
- 1(a) Percentage relative error of a balance reading5 marks
- 1(b) Partition ratio k′ and relative retention α5 marks
- 1(c) Transmittance from Beer’s law5 marks
- 1(d) Identifying a magnesium compound from its thermogram5 marks
- 1(e) Properties needed in a detector for X-ray fluorescence5 marks
- 1(f) Generation of the plasma in the ICP torch5 marks
- 1(g) Calorific value of a fuel5 marks
- 1(h) AAS and ICP compared: accuracy, interferences, detection limit5 marks