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Chemistry · Practical Organic Chemistry

JEE Main 2024 — 9 April, Shift 2 — Question 78

In the given TLC, the distance of spot A&BA \& B are 5 cm & 7 cm , from the bottom of TLC plate, respectively. RfR_{f} value of BB is x×10−1x \times 10^{-1} times more than AA. The value of xx is \qquad .

Question figure

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

Rf=Distan⁡ce   moved  by  substan⁡ce   from  base  lineDistan⁡ce  moved  by   solvent  from   base  line{{R}_{f}}=\frac{Dis\tan ce\,\,\,moved\,\,by\,\,subs\tan ce\,\,\,from\,\,base\,\,line}{Dis\tan ce\,\,moved\,\,by\,\,\,solvent\,\,from\,\,\,base\,\,line} (Rf)A=48(Rf)B=68(Rf)A(Rf)B=68×84(Rf)B=1.5(Rf)Ax=15\begin{aligned}& {{\left( {{R}_{f}} \right)}_{A}}=\frac{4}{8} \\ & {{\left( {{R}_{f}} \right)}_{B}}=\frac{6}{8} \\ & \frac{{{\left( {{R}_{f}} \right)}_{A}}}{{{\left( {{R}_{f}} \right)}_{B}}}=\frac{6}{8}\times \frac{8}{4} \\ & {{\left( {{R}_{f}} \right)}_{B}}=1.5{{\left( {{R}_{f}} \right)}_{A}} \\ & x=15 \\ \end{aligned}
Solution figure

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Chemistry
Chapter
Practical Organic Chemistry
Topic
Distillation and chromatography

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