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 __________.
Explanation
To determine the ratio of the $$\mathrm{R}_f$$ values of sample $$\mathrm{A}$$ and sample $$\mathrm{C}$$, follow these steps:
- Understanding the $$\mathrm{R}_f$$ Value:
The $$\mathrm{R}_f$$ value (Retention factor) is calculated using the formula:
$$\mathrm{R}_{\mathrm{f}} = \frac{\text{Distance travelled by the substance from the base line }(\mathrm{x})}{\text{Distance travelled by the solvent from the base line }(\mathrm{y})}$$
- 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, $$\left(R_f\right)_A = \frac{5}{12.5}$$
- 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, $$\left(R_f\right)_C = \frac{10}{12.5}$$
- 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:
$$\frac{\left(R_f\right)_A}{\left(R_f\right)_C} = \frac{5}{12.5} \times \frac{12.5}{10} = \frac{5}{10} = 0.5$$
- 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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