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Basics of Organic Chemistry question

2023 · 29 Jan · Shift 1 · Q12
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Basics of Organic Chemistry question

2023 · 29 Jan · Shift 1 · Q12

JEE MainChemistryBasics of Organic ChemistryNumerical+4 / −1
Following chromatogram was developed by adsorption of compound 'A' on a 6 cm TLC glass plate. Retardation factor of the compound 'A' is ‾×10−1\underline{\hspace{2cm}}\times 10^{-1}​×10−1. JEE Main 2023 (Online) 29th January Morning Shift Chemistry - Basics of Organic Chemistry Question 110 English
Numerical answer
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Correct answer: 6

  1. In thin layer chromatography (TLC), the retardation factor is Rf=distance travelled by solutedistance travelled by solvent frontR_f = \frac{\text{distance travelled by solute}}{\text{distance travelled by solvent front}}Rf​=distance travelled by solvent frontdistance travelled by solute​

  2. The TLC plate length is given as 6 cm6\ \text{cm}6 cm. From the chromatogram, compound AAA has moved 3.6 cm3.6\ \text{cm}3.6 cm while the solvent front has moved the full 6 cm6\ \text{cm}6 cm.

  3. Therefore, Rf=3.66=0.6R_f = \frac{3.6}{6} = 0.6Rf​=63.6​=0.6

  4. The question asks for the value in the form ‾×10−1\underline{\hspace{2cm}} \times 10^{-1}​×10−1 Since 0.6=6×10−10.6 = 6 \times 10^{-1}0.6=6×10−1

the required integer is 666.

  1. Comparison with stored answer:
  • Derived answer: 666
  • Stored correct answer: 666
  • These match.
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