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JEE Advanced levelSingle correct optionDifficulty 3 of 3si103

In a TLC experiment the solvent front moved $7.5 cm$ above the baseline. Compound X has $R_{f} = 0.60$ and compound Y has $R_{f} = 0.32$ . The distance between the centres of the two spots is

A0.28 cm
B4.5 cm
C2.1 cm
D2.4 cm

Answer: C

Distance moved by X

= 0.60 \times 7.5 = 4.50

cm and by Y

= 0.32 \times 7.5 = 2.40

cm. The separation is

4.50 - 2.40 = 2.1

cm.

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