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 = 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 acid adsorbed per mg of charcoal = 543 =18 mg

  • 47
18mg
  • -15
it is question asked in jee-mains this year and the answer is 18mg
  • -22
I wasn't able to solve this que : (
  • -26
What are you looking for?