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Question

3g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After a hour it was filtered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid absorbed (per gram of charcoal) is:

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

The correct option is D 18mg
Initial millimoles of acetic acid = molarity × volume = 50×0.06 = 3
Final millimoles of acetic acid = 50×0.042 = 2.1
Acetic acid that get adsorbed = 3 - 2.1 = 0.9 moles.
Weight of acetic acid that get adsorbed after 1 hour i.e 60 minutes = 0.9×60103 = 54 mg
Amount of acetic adsorbed per mg of charcoal = 543 = 18 mg

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