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:

Step-by-step solution

Milliequivalents ( M eqs ) of CH 3 COOH (initial) = 50 × 0.06 = 3 Milliequivalents of CH 3 COOH (final) = 50 × 0.042 = 2.1 CH 3 COOH adsorbed = 3 - 2.1 = 0.9 M eqs = 9 × 1 0 -1 × 60 g / mol × 1 0 - 3 g = 5 4 0 × 1 0 - 4 = 0.054 g = 54   mg Amount of acetic acid adsorbed per gram of charcoal = 5 4 3 = 18 mg/g of charcoal.