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

Step 1: Understand the adsorption picture.
Charcoal soaks up some acetic acid, so the solution becomes weaker. The amount adsorbed equals the amount that disappeared from solution, which we find from the drop in concentration.
Step 2: Find the equivalents present at the start.
Milliequivalents $=$ volume in mL $\times$ normality. Start: $100 \times 0.06 = 6$ meq.
Step 3: Find the equivalents left after filtering.
End: $100 \times 0.04 = 4$ meq.
Step 4: Find how much was adsorbed.
Adsorbed $= 6 - 4 = 2$ meq of acetic acid.
Step 5: Convert to mass in milligrams.
The equivalent mass of acetic acid is $60$, so $1$ meq weighs $60$ mg. Mass adsorbed $= 2 \times 60 = 120$ mg.
Step 6: Divide by the mass of charcoal.
There are $2$ g of charcoal, so per gram we get $\dfrac{120}{2} = 60$ mg/g.
\[ \boxed{60} \]