JEE Main 2015ChemistrySurface ChemistryMediumMCQ

JEE Main 2015 — Surface Chemistry Question with Solution

JEE Main 2015 (04 Apr)

Question

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour, it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

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Show full solutionCorrect option: B
Correct answer
B18 mg

Step-by-step explanation

Milliequivalents (Meqs) of CH3COOH (initial)=50×0.06=3

Milliequivalents of CH3COOH (final)=50×0.042=2.1

CH3COOH adsorbed=3-2.1=0.9 Meqs

=9×10-1×60 g/mol×10-3g

= 5 4 0 × 1 0 - 4 = 0.054 g

=54 mg

Amount of acetic acid adsorbed per gram of charcoal=543=18 mg/g of charcoal.

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About this question

This is a previous-year question from JEE Main 2015, 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.