Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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:
ChemistrySurface ChemistryJEE MainJEE Main 2015 (04 Apr)
Options:
  • A 54 mg
  • B 18 mg
  • C 36 mg
  • D 42 mg
Solution:
1851 Upvotes Verified Answer
The correct answer is: 18 mg
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.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.