JEE MAIN - Chemistry (2024 - 5th April Evening Shift - No. 23)

Using the given figure, the ratio of $$\mathrm{R}_f$$ values of sample $$\mathrm{A}$$ and sample $$\mathrm{C}$$ is $$x \times 10^{-2}$$. Value of $$x$$ is __________.

JEE Main 2024 (Online) 5th April Evening Shift Chemistry - Basics of Organic Chemistry Question 23 English

Answer
50

Explanation

To determine the ratio of the $$\mathrm{R}_f$$ values of sample $$\mathrm{A}$$ and sample $$\mathrm{C}$$, follow these steps:

  1. Understanding the $$\mathrm{R}_f$$ Value:

The $$\mathrm{R}_f$$ value (Retention factor) is calculated using the formula:

$$\mathrm{R}_{\mathrm{f}} = \frac{\text{Distance travelled by the substance from the base line }(\mathrm{x})}{\text{Distance travelled by the solvent from the base line }(\mathrm{y})}$$

  1. Calculate $$\mathrm{R}_f$$ for Sample $$A$$:
  • Distance travelled by substance $$A$$ (from the figure): 5 cm

  • Distance travelled by the solvent: 12.5 cm

  • So, $$\left(R_f\right)_A = \frac{5}{12.5}$$
  1. Calculate $$\mathrm{R}_f$$ for Sample $$C$$:
  • Distance travelled by substance $$C$$ (from the figure): 10 cm

  • Distance travelled by the solvent: 12.5 cm

  • So, $$\left(R_f\right)_C = \frac{10}{12.5}$$
  1. Ratio of $$\mathrm{R}_f$$ values for $$A$$ and $$C$$:
  • The ratio is given by:

$$\frac{\left(R_f\right)_A}{\left(R_f\right)_C} = \frac{\frac{5}{12.5}}{\frac{10}{12.5}}$$

  • Simplify the ratio:

$$\frac{\left(R_f\right)_A}{\left(R_f\right)_C} = \frac{5}{12.5} \times \frac{12.5}{10} = \frac{5}{10} = 0.5$$

  1. Converting to the given form:
  • The simplified ratio (0.5) should be expressed in the form $$x \times 10^{-2}$$.

  • Since $$0.5 = 50 \times 10^{-2}$$,

we can identify $$x = 50$$.

Therefore, the value of $$x$$ is 50.

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