`3g` of actived chacoal was added to 50mL 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. 42 mg
B. 54 mg
C. 18 mg
D. 36 mg

Answer

C

Explanation

Let the weight ofacetic acid initially be `w_(1)` in 50 mL of 0.060 N solution.
Let the `N= (w_(1) xx 1000)/(M.wt. xx 50)` (Normality =0.06 N)
`0.06=(w_(1) xx 1000)/(60 xx 50)`
`implies w_(1)=(0.06 xx 60 xx 50)/(1000) =0.18 g = 180 mg`
After an hour, the strength of acetic acid = 0.042 N so, let the weight of acetic acid be `w_(2)`
 `N= (w_(2) xx 1000)/(60 xx 50)`
`0.042=(w_(2) xx 1000)/(3000)`
`implies w_(2)=0.126 g=126 mg`
So amount of acetic acid adsorbed per 3g
= 180-126 mg = 54 mg
`:.` amount of acctic acid absorbed per `g = 54 // 3 = 18g`