JEE MAIN - Chemistry (2024 - 9th April Evening Shift - No. 22)
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In the given TLC, the distance of spot A & B are 5 cm & 7 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 __________.
Answer
15
Explanation
$$\mathrm{R}_{\mathrm{f}}=\frac{\text { Distance moved by substance from base line }}{\text { Distance moved by solvent from base line }}$$
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$$\begin{aligned} & \left(\mathrm{R}_{\mathrm{f}}\right)_A=\frac{4}{8} \quad\left(\mathrm{R}_{\mathrm{f}}\right)_B=\frac{6}{8} \\ & \frac{\left(\mathrm{R}_{\mathrm{f}}\right)_B}{\left(\mathrm{R}_{\mathrm{f}}\right)_A}=\frac{6}{8} \times \frac{8}{4} \\ & \left(\mathrm{R}_{\mathrm{f}}\right)_B=1.5\left(\mathrm{R}_{\mathrm{f}}\right)_A \\ & x=15 \end{aligned}$$
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