JEE Main 2021ChemistrySurface ChemistryHardNumerical

JEE Main 2021 — Surface Chemistry Question with Solution

JEE Main 2021 (31 Aug Shift 2)

Question

CH4 is adsorbed on 1 g charcoal at 0°C following the Freundlich adsorption isotherm. 10.0 mL of CH4 is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH4 adsorbed at 300 mm of Hg is 10x mL. The value of x is _________ ×10-2. (Nearest integer)
[Use log102=0.3010, log103=0.4771]

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Show full solutionCorrect answer: 128
Correct answer
128

Step-by-step explanation

xm=Kp1n

x∝V for the same gas
101=K1001n   ...(1)
151=K2001n   ....(2)
V1=K3001n   ...(3)
Divide (2) to (1) =1510=21n
log32=1nlog2
1n=0.4771-0.30100.3010=0.585
Divide (3) to (1) =V10=31n
logV10=1nlog3
logV10=0.585×0.4771=0.2791
V10=100.2791
V=10×100.2791=101.2791=10x
x=1.2791=127.91×10-2≈128×10-2

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About this question

This is a previous-year question from JEE Main 2021, covering the Surface Chemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.