# 3g activated charcoal was added to 50ml of acetic acid solution 0.06N in a flask after an hour it was filtered & the strength of the filteratewas found to be 0.042N the amount of acetic acid adsorbed (per gram of charcoal )

Initial millimoles of acetic acid = molarity $×$ volume =
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  =
Amount of acetic acid adsorbed per mg of charcoal =

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18mg
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it is question asked in jee-mains this year and the answer is 18mg
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I wasn't able to solve this que : (
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