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

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: $x = 15$.

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: $x = 15$.