3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour, It was filtered and the strength of the filtrate was found to be 0.42 N. Calculate the amount of acetic acid adsorbed per gram of charcoal.

Explanation

Amount of acetic acid initially present
Molarity of `CH_(3)COOH` solution `=("Mass of acetic acid/Molar mass")/("Volume of solution in liter")" " ("Acetic acid is a monobasic acid")`
`(0.06mol L^(-1))=(W)/((60g mol^(-1))xx(0.05L))`
`W=(0.042mol L^(-1))xx(60g mol^(-1))xx(0.05L)=0.126g`
Amount of acetic acid actually adsorbed = (0.180-0.126)=0.054g = 54 mg
Amount of charcoal available = 3.0 g
`therefore` Amount of acetic acid adsorbed per gram of charcoal `=((54mg))/((3.0g))xx(1.0g)=18mg`