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

Correct answer: 0 R_ f = Distance moved by substance from base line Distance moved by solvent from base line aligned & (R_f)_A=(4)/(8) (R_f)_B=(6)/(8) & ((R_f)_B)/((R_f)_A)=(6)/(8) × (8)/(4) & (R_f)_B=1.5(R_f)_A & x=15 aligned

JEE Main 20249 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Evening Shift
Subject: Chemistry
Chapter: General Organic Chemistry

JEE Main 2024 · 9 April · Evening Shift · Chemistry · General Organic Chemistry

In the given TLC, the distance of spot A & B are (5 ~cm ~\&~ 7 ~cm...

$In the given TLC, the distance of spot A & B are (5 ~cm ~\&~ 7 ~cm ), from the bottom of TLC plate, respectively. ( R_$
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 ______.

Correct answer

Step-by-step solution

\mathrm{R}_{\mathrm{f}}=\frac{\text{ Distance moved by substance\text{ 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}

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