3 g of activated charcoal was added to 50 mL of acetic acid solution
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:
- 54 mg
- 42 mg
- 36 mg
- 18 mg
Answer
Amount adsorbed
= (0.060 – 0.042) x 50 x 10–3 x 60
= 0.018 x 50 x 60 x 10–3
= 0.018 x 3
= 0.054 gm
= 54 mg
Amount adsorbed per gram of activated charcoal = 54/3 = 18 mg
The correct option is D.