Chemistry · Solid State

JEE Main 2024 — 5 April, Shift 2 — Question 83

Using the given figure, the ratio of RfR_{f} values of sample AA and sample CC is x×10−2\mathrm{x} \times 10^{-2}. Value of x is \qquad . Samples (A,B,C)

figure

Fig : Paper chromatography of Samples

Answer: 50

Numerical answer — enter this value.

Step-by-step solution

Rf\mathrm{R}_{\mathrm{f}} of A=512.5Rf\mathrm{A}=\frac{5}{12.5} \quad \mathrm{R}_{\mathrm{f}} of C=1012.5\mathrm{C}=\frac{10}{12.5}

Ratio =Rf(A)Rf(C)=12=0.5=\frac{\mathrm{R}_{\mathrm{f}(\mathrm{A})}}{\mathrm{R}_{\mathrm{f}(\mathrm{C})}}=\frac{1}{2}=0.5 or

50×10−250 \times 10^{-2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
Solid State
Topic
Distillation and chromatography
Using the given figure, the ratio of R f values of sample A and… | JEE Main 2024 PYQ with Solution · DhiX AI