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

Question

Aicharya · Accepted Answer

$\mathrm{R}_{\mathrm{f}}=\frac{	ext { Distance moved by substance from base line }}{	ext { Distance moved by solvent from base line }}$ $\begin{aligned} & \left(R_fight)_A=\frac{4}{8} \quad\left(R_fight)_B=\frac{6}{8} \ & \frac{\left(R_fight)_B}{\left(R_fight)_A}=\frac{6}{8} 	imes \frac{8}{4} \ & \left(R_fight)_B=1.5\left(R_fight)_A \ & x=15\end{aligned}$

In the given TLC, the distance of spot A & B are $5 \mathrm{~cm}$ & $7 \mathrm{~cm}$, from the bottom of TLC…

Solution