Spectrophotometry and Beer's Law: Quiz

7 questions

  1. Question 1EasyA solution transmits 50.0 % of the light. What is its absorbance?
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    Answer: 0.301

    T = 0.500, so A = −log(0.500) = 0.301. Using 50.0 instead of 0.500 gives −1.70, which is impossible for a real sample.

  2. Question 2EasyIf the concentration of a solution is doubled (same cell), the absorbance:
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    Answer: doubles

    Beer's law, A = εbc: A is directly proportional to c.

  3. Question 3MediumWhat are the units of molar absorptivity, ε?
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    Answer: L mol⁻¹ cm⁻¹

    A has no unit, so ε = A ÷ (b × c) has units 1 ÷ (cm × mol/L) = L mol⁻¹ cm⁻¹.

  4. Question 4MediumA = 0.450, ε = 1.50 × 10⁴ L mol⁻¹ cm⁻¹, b = 1.00 cm. What is c?
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    Answer: 3.00 × 10⁻⁵ mol/L

    c = A ÷ (εb) = 0.450 ÷ (1.50 × 10⁴ L mol⁻¹ cm⁻¹ × 1.00 cm) = 3.00 × 10⁻⁵ mol/L.

  5. Question 5EasyWhat is the purpose of the blank?
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    Answer: to zero the instrument so that only the analyte absorbance is measured

    The blank contains everything except the analyte, so absorption by the solvent, reagents and cell is subtracted.

  6. Question 6MediumThe same solution is moved from a 1.00 cm cell to a 2.00 cm cell. Its absorbance changes from 0.250 to:
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    Answer: 0.500

    A is proportional to the path length b: doubling b doubles A.

  7. Question 7HardAn unknown gives A = 1.85, but the standards only go up to A = 0.95. What should you do?
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    Answer: dilute the unknown by a known factor, measure again and multiply by the factor

    The reading must fall inside the calibration range; Beer's law may not hold at high absorbance. A longer cell would raise A even further.