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JEE Chemistry 2024 Question with Solution

On a thin layer chromatographic plate, an organic compound moved by 3.5 cm3.5 \, \text{cm}, while the solvent moved by 5 cm5 \, \text{cm}. The retardation factor of the organic compound is ×10−1\times 10^{-1}:

Answer

Correct answer:7

Step-by-step solution

Standard Method

Given: The organic compound moved by 3.5 cm3.5 \, \text{cm} and the solvent front moved by 5 cm5 \, \text{cm}.

Find: The retardation factor in the form ____×10−1\_\_\_\_ \times 10^{-1}.

In thin layer chromatography, the retardation factor is given by

Rf=Distance travelled by the compoundDistance travelled by the solvent frontR_f = \frac{\text{Distance travelled by the compound}}{\text{Distance travelled by the solvent front}}

Substituting the given values,

Rf=3.55.0=0.7R_f = \frac{3.5}{5.0} = 0.7

Now express 0.70.7 in the required form:

0.7=7×10−10.7 = 7 \times 10^{-1}

Therefore, the required numerical value is 77.

Common mistakes

  • Using the inverse ratio, solvent distancecompound distance\frac{\text{solvent distance}}{\text{compound distance}}, is incorrect because RfR_f is defined as distance travelled by the compound divided by distance travelled by the solvent front. Always place the compound distance in the numerator.

  • Reporting the answer as 0.70.7 instead of 77 is incorrect here because the question asks for the value in the form ____×10−1\_\_\_\_ \times 10^{-1}. Convert 0.70.7 to 7×10−17 \times 10^{-1} before writing the final numerical answer.

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