See image | AITS & Test Series — Chem-Mantra

AITS & Test SerieshardMCQ SINGLE

Question

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Answer: C

💡 Solution & Explanation

1/n x P m  x 1 log logK logP m n   o 1 tan30 P 1atm n   x log logK m  As log K = 0.477 x 3g m  per 1 g of adsorbent

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