JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 15)
Using the provided information in the following paper chromatogram :
_25th_February_Morning_Shift_en_15_1.png)
Fig : Paper chromatography for compounds A and B.
the calculated Rf value of A ________$$\times$$ 10-1.
Fig : Paper chromatography for compounds A and B.
the calculated Rf value of A ________$$\times$$ 10-1.
Answer
4
Explanation
$${R_f} = {{Dis\tan ce\,travelled\,by\,compound} \over {Dis\tan ce\,travelled\,by\,solvent}}$$
On chromatogram distance travelled by compound is = 2 cm
Distance travelled by solvent = 5 cm
So, $${R_f} = {2 \over 5} = $$ 4 $$\times$$ 10$$-$$1 = 0.4
On chromatogram distance travelled by compound is = 2 cm
Distance travelled by solvent = 5 cm
So, $${R_f} = {2 \over 5} = $$ 4 $$\times$$ 10$$-$$1 = 0.4
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