JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization
The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance \(A\) and substance \(B\) from the base line are \(3.25\,cm .2 .08\) \(cm\) and \(1.05\,cm\). respectively. The ratio of \(R_{f}\) values of \(A\) to \(B\) is\(.........\)
- A \(5\)
- B \(3\)
- C \(8\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(\frac{ R _{ F _{ A }}}{ R _{ F _{ B }}}=\frac{\frac{2.08}{3.25}}{\frac{1.05}{3.25}}=\frac{2.08}{1.05} \simeq 2\)
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