Question
$$3 g$$ of activated charcoal was added to $$50 mL$$ of acetic acid solution $$(0.06N)$$ in a flask. After an hour it was filtered and the strength of the filtrate was found to be $$0.042 N.$$ The amount of acetic acid adsorbed ( per gram of charcoal ) is :
A.
42 $$mg$$
B.
54 $$mg$$
C.
18 $$mg$$
D.
36 $$mg$$
Answer :
18 $$mg$$
Solution :
Let the weight of acetic acid initially be w1 in 50 ml of 0.060 N solution.
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,N = \frac{{{w_1} \times 1000}}{{M.wt. \times 50}}\,\,\,\,\,\,\,\,\left( {{\text{Normality}} = 0.06\,N} \right) \cr
& 0.06 = \frac{{{w_1} \times 1000}}{{60 \times 50}} \cr
& \Rightarrow \,\,\,{w_1} = \frac{{0.06 \times 60 \times 50}}{{1000}} \cr
& = 0.18\,g \cr
& = 180\,mg. \cr} $$
After an hour, the strength of acetic acid = 0.042 N
so, let the weight of acetic acid be w2
$$\eqalign{
& N = \frac{{{w_2} \times 1000}}{{60 \times 50}} \cr
& 0.042 = \frac{{{w_2} \times 1000}}{{3000}} \cr
& \Rightarrow {w_2} = 0.126\,g \cr
& = 126\,mg \cr} $$
So amount of acetic acid adsorbed per 3g
= 180 -126 $$mg$$ = 54 $$mg$$
Amount of acetic acid adsorbed per g
$$ = \frac{{54}}{3} = 18mg$$