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AITS & Test SerieshardMCQ SINGLE

Question

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Answer: A

💡 Solution & Explanation

x 1 log logk logP m n   Log k = 0.699  k = 5 Slope = tan = tan 45° = 1 = 1 n  1 n x k.p m  = 5  (0.5)1/1 = 2.5 g/g adsorbent.

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