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 (2015 JEE Main)

Key Concepts Needed

1. Normality (N): For monoprotic acids like acetic acid, 1 mole = 1 equivalent.
2. Equivalents or moles present:
   $\text{Equivalents} = N \times V_{\text{(in L)}}$
3. Adsorption: Activated charcoal can hold (adsorb) molecules onto its surface, decreasing their concentration in solution.
4. Mass from moles:
   $\text{Mass} = n \times M_{\text{molar}}$
   For acetic acid, $M_{\text{molar}} = 60\,\text{g mol}^{-1}$.

Logical Chain a Student Should Follow

1. Initial Content – Compute the initial equivalents (same as moles here) of acetic acid in 50 mL at 0.06 N.
2. Remaining Content – Compute the equivalents left after adsorption when normality drops to 0.042 N.
3. Difference – Subtract to get the equivalents (moles) adsorbed by charcoal.
4. Convert to Mass – Multiply moles adsorbed by molar mass of acetic acid to know grams adsorbed.
5. Per-Gram Basis – Divide that mass by the mass of charcoal used (3 g) to find grams per gram.

Each step flows directly from a standard stoichiometry calculation to a simple division, matching the definition of adsorption capacity.

🎈 What is happening here?

Imagine you have a sponge (charcoal) that can soak up vinegar (acetic acid) from water. You start with a glass that has a known amount of vinegar in it. After letting the sponge sit in the glass for an hour, you notice the water now has less vinegar because the sponge soaked some of it up.

🍭 What do we need to find?

We want to know how much vinegar, by weight, each gram of the sponge soaked up.

🛠️ How will we do it, in kid-friendly steps?

1. Count how much vinegar was there before using the strength (0.06 N) and the size of the glass (50 mL).
2. Count how much vinegar is left after the sponge sat for an hour (0.042 N in the same 50 mL).
3. Find the difference: that is the vinegar soaked up.
4. Change the soaked-up amount from moles to grams using the weight of 1 mole of vinegar.
5. Divide by how many grams of sponge we used to get how much vinegar 1 gram of sponge soaked.

That’s it! 🥳

Step-by-Step Calculation

1. Initial equivalents (moles) of acetic acid
   Volume $V = 50\,\text{mL} = 0.05\,\text{L}$
   Normality $N_{\text{initial}} = 0.06$

   $\text{Equivalents}_{\text{initial}} = 0.06 \times 0.05 = 0.003$

2. Equivalents after adsorption
   $N_{\text{final}} = 0.042$

   $\text{Equivalents}_{\text{final}} = 0.042 \times 0.05 = 0.0021$

3. Equivalents (moles) adsorbed

   $n_{\text{adsorbed}} = 0.003 - 0.0021 = 0.0009\,\text{mol}$

4. Mass of acetic acid adsorbed
   Molar mass $M = 60\,\text{g mol}^{-1}$

   $\text{Mass}_{\text{adsorbed}} = 0.0009 \times 60 = 0.054\,\text{g}$

5. Adsorption per gram of charcoal
   Charcoal used $m_{\text{char}} = 3\,\text{g}$

   $\text{Adsorption capacity} = \frac{0.054}{3} = 0.018\,\text{g g}^{-1}$

Final Answer

$0.018\,\text{g of acetic acid per gram of activated charcoal}$
(Equivalent to 18 mg g⁻¹)