Verified Solution • 2025–26 Session moderate

QJEEM21Cha844: JEE Main 2021 Chemistry - Class 12 JEEM21Cha844 Solution: CH 4 is absorbed on 1g charcoal at 0 C following the Freundlich …

CH4 is absorbed on 1g charcoal at 0 ∘^\circ C following the Freundlich adsorpiton 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 _________ ×\times 10-2.

(Nearest integer) [Use log102 = 0.3010, log103 = 0.4771]

Step-by-Step Verified Solution

We know

xm=KP1/n{x \over m} = K{P^{1/n}} ; using (x ∝\propto V)

⇒101=K×(100)1/n\Rightarrow {{10} \over 1} = K \times {(100)^{1/n}} ..... (1)

151=K×(200)1/n{{15} \over 1} = K \times {(200)^{1/n}} ..... (2)

V1=K×(300)1/n{V \over 1} = K \times {(300)^{1/n}} ..... (3)

Divide

(2) / (1)

1510=21/n{{15} \over {10}} = {2^{1/n}}

log⁡(32)=1nlog⁡2\log \left( {{3 \over 2}} \right) = {1 \over n}\log 2

1n=log⁡3−log2log⁡2=0.4771−0.30100.3010{1 \over n} = {{\log 3 - log2} \over {\log 2}} = {{0.4771 - 0.3010} \over {0.3010}}

1n=0.585{1 \over n} = 0.585

Divide

(3) / (1)

V10=31/n{V \over {10}} = {3^{1/n}}

log⁡(V10)=1nlog⁡3\log \left( {{V \over {10}}} \right) = {1 \over n}\log 3

log⁡(V10)=0.585×0.4771=0.2791\log \left( {{V \over {10}}} \right) = 0.585 \times 0.4771 = 0.2791

V10=100.279⇒V=10×100.279{V \over {10}} = {10^{0.279}} \Rightarrow V = 10 \times {10^{0.279}}

⇒V=101.279=10x\Rightarrow V = {10^{1.279}} = {10^x}

⇒\Rightarrow x = 1.279

⇒\Rightarrow x = 128 ×\times 10 −- 2 (Nearest integer)

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring