Question

In a flask, 2 g of activated charcoal was added to 100 mL of acetic acid solution of 0.06 N. After 2 hours, the solution was filtered. The concentration of filtrate was found to be 0.04 N. The mass of acetic acid (in mg) adsorbed per gram of charcoal is:

• A. 30
• B. 60
• C. 90
• D. 120

Hint:

Ensure all units are converted properly (e.g., meq to mg) when calculating mass adsorbed per unit mass of adsorbent.

Answer 1

The Correct Option is B

Concept: Adsorption amount is calculated by the difference in concentration before and after the adsorption process multiplied by the volume.

Step 1: Calculate the moles/equivalents adsorbed.
Initial equivalents = \(100 \text{ mL} \times 0.06 \text{ N} = 6 \text{ meq}\)
Final equivalents = \(100 \text{ mL} \times 0.04 \text{ N} = 4 \text{ meq}\)
Equivalents adsorbed = \(6 - 4 = 2 \text{ meq}\)

Step 2: Convert equivalents to mass (mg).
Equivalent mass of acetic acid (\(CH_3COOH\)) = \(60 \text{ g/eq} = 60 \text{ mg/meq}\).
Mass of acetic acid adsorbed = \(2 \text{ meq} \times 60 \text{ mg/meq} = 120 \text{ mg}\).

Step 3: Calculate adsorption per gram of charcoal.
Total charcoal = 2 g. Adsorption per gram = \(\frac{120 \text{ mg}}{2 \text{ g}} = 60 \text{ mg/g}\). \[ \boxed{60} \]