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In the given TLC, the distance of spot A & B are \(5 \mathrm{~cm} ~\&~ 7 \mathrm{~cm}\), from the bottom of TLC plate, respectively.
\(\mathrm{R}_{\mathrm{f}}\) value of \(\mathrm{B}\) is \(x \times 10^{-1}\) times more than \(\mathrm{A}\). The value of \(x\) is ______.
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Verified Answer
The correct answer is:
15
$\mathrm{R}_{\mathrm{f}}=\frac{\text { Distance moved by substance from base line }}{\text { Distance moved by solvent from base line }}$
$\begin{aligned} & \left(R_f\right)_A=\frac{4}{8} \quad\left(R_f\right)_B=\frac{6}{8} \\ & \frac{\left(R_f\right)_B}{\left(R_f\right)_A}=\frac{6}{8} \times \frac{8}{4} \\ & \left(R_f\right)_B=1.5\left(R_f\right)_A \\ & x=15\end{aligned}$
$\begin{aligned} & \left(R_f\right)_A=\frac{4}{8} \quad\left(R_f\right)_B=\frac{6}{8} \\ & \frac{\left(R_f\right)_B}{\left(R_f\right)_A}=\frac{6}{8} \times \frac{8}{4} \\ & \left(R_f\right)_B=1.5\left(R_f\right)_A \\ & x=15\end{aligned}$
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