Chapter 11: Problem 196
\(3 \mathrm{~g}\) of activated charcoal was added to \(50 \mathrm{~mL}\) of acetic acid solution \((0.06 \mathrm{~N})\) in a flask. After an hou it was filtered and the strength of the filtrate was found to be \(0.042 \mathrm{~N}\). The amount of acetic acid adsorbed (per gram of charcoal) is (a) \(18 \mathrm{mg}\) (b) \(36 \mathrm{mg}\) (c) \(42 \mathrm{mg}\) (d) \(54 \mathrm{mg}\)
Short Answer
Step by step solution
Initial Moles of Acetic Acid
Moles of Acetic Acid in Filtrate
Moles of Acetic Acid Adsorbed
Convert Moles to Mass
Mass Adsorbed per Gram of Charcoal
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acetic Acid
Activated Charcoal
Molar Mass Calculation
- Carbon (C): 12.01 g/mol, and there are two carbons.
- Hydrogen (H): 1.01 g/mol, and there are four hydrogens.
- Oxygen (O): 16.00 g/mol, and there are two oxygens.