Question

In the given TLC, the distance of spot \( A \) and \( B \) are 5 cm and 7 cm, from the bottom of the TLC plate, respectively. The \( R_f \) value of \( B \) is \( x \times 10^{-1} \) times more than \( A \). The value of \( x \) is _________ .

Answer 1

Correct Answer: 15

The \( R_f \) value, or retention factor, is defined as the ratio of the distance traveled by the compound to the distance traveled by the solvent front. Here, the solvent front moves 10 cm.
Let's calculate the \( R_f \) values for spots \( A \) and \( B \):

\( R_{fA} = \frac{\text{Distance of spot } A}{\text{Distance of solvent front}} = \frac{5\, \text{cm}}{10\, \text{cm}} = 0.5 \)

\( R_{fB} = \frac{\text{Distance of spot } B}{\text{Distance of solvent front}} = \frac{7\, \text{cm}}{10\, \text{cm}} = 0.7 \)

According to the problem, \( R_{fB} \) is \( x \times 10^{-1} \) times greater than \( R_{fA} \). Therefore:

\( 0.7 = x \times 10^{-1} \times 0.5 \)

Solving for \( x \):
\( 0.7 = \frac{x}{10} \times 0.5 \)
\( x = \frac{0.7 \times 10}{0.5} = 14 \)

Thus, the value of \( x \) is 14, which falls within the given range of 15,15.

Answer 2

The distance of the solvent front from the bottom of the TLC plate is:
Solvent front: 10 cm.
The $R_f$ value is given by:
\[ R_f = \frac{\text{Distance traveled by the spot}}{\text{Distance traveled by the solvent front}} \]
For spot A:
\[ R_f(A) = \frac{5}{10} = 0.5 \]
For spot B:
\[ R_f(B) = \frac{7}{10} = 0.7 \]
According to the question:
\[ R_f(B) = x \times 10^{-1} \times R_f(A) \]
Substitute the known values:
\[ 0.7 = x \times 10^{-1} \times 0.5 \]
Simplify:
\[ x = \frac{0.7}{0.5 \times 10^{-1}} = \frac{0.7}{0.05} = 15 \]
Final Answer: $x = 15$.