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

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Practical and Analytical Chemistry

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

x \times 1 0^{- 1}

times more than

A

. The value of

x

is ____.

Answer: 15

Solution

R_{f} = \frac{distance travelled by spot}{distance travelled by solvent front}

From the plate, spot A and spot B are 4 cm and 6 cm from the baseline respectively, with the solvent front 8 cm from the baseline:

R_{f, A} = \frac{4}{8} = 0.5, R_{f, B} = \frac{6}{8} = 0.75

\frac{R _{f, B}}{R _{f, A}} = \frac{0.75}{0.5} = 1.5 = 15 \times 1 0^{- 1}

x = 15

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer

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