Question

3 g of activated charcoal was added to 50 mL of the 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:
A. 18 mg
B. 36 mg
C. 42 mg
D. 54 mg

Answer 1

Activated charcoal has a property to adsorb acetic acid molecules on its surface. To calculate the acetic acid adsorbed on the surface of the charcoal in mill moles we should know the acetic concentration in millimoles before the reaction and after the reaction.

Complete step by step answer:
- We have to calculate the amount of acetic acid adsorbed on the surface of the activated charcoal after one hour.
- The concentration of the acetic acid is 0.06 N.
- Final concentration of the acetic acid left after adsorption is 0.042 N.
- The volume of the acetic acid is 50 mL.
- milli moles of acetic acid before adsorption = (0.06) (50) = 3
- milli moles of acetic acid after adsorption = (0.042) (50) = 2.1
- Therefore milli moles of acetic acid adsorbed = 2- 2.1 = 0.9 milli moles.
- Weight of acetic acid adsorbed on the charcoal surface after 1 hour (60 minutes) is as follows.

& = \dfrac{0.9\times 60}{1000} \\\ & = 54mg \\\ \end{aligned}$$ \- 54 mg on 3 grams of charcoal, but we have to calculate for one gram as per the question. \- Therefore $\dfrac{54}{3}=18 mg$ . **So, the correct answer is “Option A”.** **Note:** The activated charcoal contains an active surface, it has a capability to hold some chemicals on its surface through a process called adsorption. Adsorption is a surface phenomenon. Means the chemicals are going to adsorb means the chemical won’t react. They stay on the surface only.