Discussion Forum

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 0.042 N. The amount of acetic acid adsorbed (per  gram of charcoal ) is

Answer:

Explanation:

Given, initial strength od acetic acid =0.06 N

 Final strength =0.042 N

Volume given  =50mL

 $\therefore$  Initial m moles of   $CH_{3}COOH$

 $=0.06\times50=3$

 Final m moles of  $CH_{3}COOH$

 $=0.042\times50=2.1$

 $\therefore$     m moles of $CH_{3}COOH$ adsorbed =3-2.1  =0.9m mol

 Hence , mass of $CH_{3}COOH$ absorbed per gram of charcoal

                                         $=\frac{0.9\times60}{3}$

  (   $\therefore$ molar mass of  $CH_{3}COOH$  = 60gmol^-1 )

                           $=\frac{54}{3}=18 mg$