(JEE Advanced 2014) of activated charcoal was added to of acetic acid solution () in a flask. After — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
(JEE Advanced 2014) $3 \text{ g}$ of activated charcoal was added to $50 \text{ mL}$ of acetic acid solution ($0.06 \text{ N}$) in a flask. After adsorption, the final concentration was $0.042 \text{ N}$. The amount of acetic acid adsorbed per gram of charcoal is:
Answer: A
💡 Solution & Explanation
$\text{Initial mEq} = 50 \times 0.06 = 3$. $\text{Final mEq} = 50 \times 0.042 = 2.1$. $\text{Adsorbed mEq} = 0.9$. $\text{Mass adsorbed} = 0.9 \times 60 \text{ (MW/1)} = 54 \text{ mg}$. $\text{Per gram} = 54 / 3 = 18 \text{ mg}$. [12]
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