QJEEM24Cha66a: JEE Main 2024 Chemistry - Class 11 JEEM24Cha66a Solution: Using the given figure, the ratio of values of sample and…

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.