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

Step 1: Find the Distance Traveled by the Solvent Front

From the provided TLC diagram:

• Total height of the plate: $$10\text{ cm}$$
• Distance from bottom edge to baseline: $$1\text{ cm}$$
• Distance from top edge to solvent front: $$1\text{ cm}$$

The total distance traveled by the solvent from the baseline is:

$$\text{Distance of solvent front} = 10\text{ cm} - 1\text{ cm} - 1\text{ cm} = 8\text{ cm}$$

Step 2: Calculate the $$R_f$$ Values for Spots A and B

The $$R_f$$ value is calculated as the distance traveled by the compound from the baseline divided by the distance traveled by the solvent front from the baseline.

• For Spot A:

  Distance from bottom edge = $$5\text{ cm}$$, so distance from baseline = $$5\text{ cm} - 1\text{ cm} = 4\text{ cm}$$.

  $$R_f(A) = \frac{4\text{ cm}}{8\text{ cm}} = 0.5$$



• For Spot B:

  Distance from bottom edge = $$7\text{ cm}$$, so distance from baseline = $$7\text{ cm} - 1\text{ cm} = 6\text{ cm}$$.

  $$R_f(B) = \frac{6\text{ cm}}{8\text{ cm}} = 0.75$$

Step 3: Solve for $$x$$

The problem states that the $$R_f$$ value of B is a multiple of A expressed as:

$$R_f(B) = (x \times 10^{-1}) \times R_f(A)$$

Substitute the calculated $$R_f$$ values into the relation:

$$0.75 = (x \times 10^{-1}) \times 0.5$$

$$x \times 10^{-1} = \frac{0.75}{0.5} = 1.5$$

$$x = 1.5 \times 10 = 15$$

Conclusion:

Comparing the relative positions shows that the $$R_f$$ value of B is exactly $$1.5$$ times that of A, which gives a value of $$15$$ for $$x$$.

Answer: 15