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QJEEM24Ch5461: JEE Main 2024 Chemistry - Class 11 JEEM24Ch5461 Solution: In the given TLC, the distance of spot A & B are 5 cm & 7 cm,…

JEE Main 2024 (Online) 9th April Evening Shift Chemistry - Practical Organic Chemistry Question 8 English

In the given TLC, the distance of spot A & B are 5 cm & 7 cm, from the bottom of TLC plate, respectively.

Rf\mathrm{R}_{\mathrm{f}} value of B\mathrm{B} is x×10−1x \times 10^{-1} times more than A\mathrm{A} . The value of xx is __________.

Step-by-Step Verified Solution

Rf= Distance moved by substance from base line  Distance moved by solvent from base line \mathrm{R}_{\mathrm{f}}=\frac{\text { Distance moved by substance from base line }}{\text { Distance moved by solvent from base line }}

JEE Main 2024 (Online) 9th April Evening Shift Chemistry - Practical Organic Chemistry Question 8 English Explanation

amp;(Rf)A=48(Rf)B=68amp;(Rf)B(Rf)A=68×84amp;(Rf)B=1.5(Rf)Aamp;x=15\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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