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 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: $x = 15$.