The volume of gases $H_{2}, C H_{4}, C O_{2}$ and $N H_{3}$ absorbed by 1 gm of activated charchol at 298 K are in the order:

H_{2} > C H_{4} > C O_{2} > N H_{3}

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C H_{4} > C O_{2} > N H_{3} > H_{2}

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C O_{2} > N H_{3} > H_{2} > C H_{4}

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N H_{3} > C O_{2} > C H_{4} > H_{2}

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Solution

The correct option is D $N H_{3} > C O_{2} > C H_{4} > H_{2}$
(D) $N H_{3} > C O_{2} > C H_{4} > H_{2}$
Explanation: Because $H_{2} = 4.7$ cc/gm, $C H_{4} = 16.2$ cc/gm, $C O_{2} = 48$ cc/gm, $N H_{3}$ is $181$ cm/gm and more over $N H_{3}$ due to hydrogen bonding can easily get liquified and hence has high volume of gas absorbed.
Option D is correct

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