Question bank/JEE Main/Chemistry

JEE MainChemistryModerateSurface Chemistry

3 gram of activated charcoal was added to 50 mL of acetic acid
solution (0.06N) 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

Show answer and solution
Initial mmoles of CH3COOH = 0.06 × 50
Final mmoles of CH3COOH = 0.042 × 50
Hence, mass of CH3COOH adsorbed per gram of charcoal
=(0.06−0.042)×50×10−3×60×1033=18mg= \frac{(0.06 - 0.042) \times 50 \times 10^{- 3} \times 60 \times 10^{3}}{3} = 18mg

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx

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