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.

Options

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

Explanation

Moles of acetic acid initially present =`(0.06xx50)/1000`
`=3.0xx 10^(-3)`
Moles of acetic acid after adsorption = `(0.042xx50)/1000`
`=2.1 xx 10^(-3)`
Moles of acetic acid adsorbed = `3.0 xx 10^(-3) -2.1 xx 10^(-3)`
`=0.9 xx 10^(-3)` moles
Moles Mass of acetic acid adsorbed `= 0.9 xx 10^(-3) xx60`
`= 54 xx 10^(-3) g`
Amount of acetic acid adsorbed per gram of charcoal
`=(54xx10^6)/3 = 18 xx 10^(-3) g = 18 mg`