Question

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 )

Answer 1

Initial millimoles of acetic acid = molarity $\times$ volume = $50\times 0.06=3$
Final millimoles of acetic acid = 50 $\times$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 = $\frac{0.9\times 60}{{10}^3}=54mg$
Amount of acetic acid adsorbed per mg of charcoal = $\frac{54}{3}=18mg$