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Question

3 g of activated charcoal was added to 50mL 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 0.042N. The amount of acetic acid adsorbed(per gram of charcoal) is?

A
18 mg
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B
36 mg
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C
42 mg
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D
54 mg
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Solution

The correct option is C 18 mg
Initial millimole of acetic acid solution = N × Volume = 0.06N×50=3 millimole
Final millimole of acetic acid solution = 0.042N×50=2.1 millimole
Acetic acid adsorbed = 3- 2.1 = 0.9 millimole
Weight after 60 minutes = 0.9103×60=54mg
Amount of acetic acid adsorbed per gram of charcoal = 543=18 mg

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