`3g` of actived chacoal was added to some of acetic acid solution `(0.06N)` in a flask. After an hour it was filterred and the strength of the filtrate was found to be `0.042N` . The amount of acetic adsorbed (per gram of charcoal) is:

Options

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

Answer

C

Explanation

Amoount adsobed is equal to number of equivalents adsorbed multiplied by gram equivalent mass:
`=(0.060N-0.042N)(50mL)((10^(-3)L)/(1mL))((60g)/(eq))`
`=0.054gm`
`=54mg(1g=10^(3)mg)`
Note thus acetic acid is monobasic, thus, its equivalent man is same as molecular mass.