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EEJ MAIN Chemistry QUESTION #7519
Question 1

3 g of activated charcoal was added to 50 mL of 0.06 N acetic acid. After filtration, the filtrate had a normality of 0.042 N. How much acetic acid was adsorbed per gram of charcoal?

  • 18 mg✔️
  • 36 mg
  • 42 mg
  • 54 mg
Correct Answer Explanation

Initial moles of acetic acid: $0.06 \times 0.050 = 3 \times 10^{-3}\ \text{mol}$

Remaining moles: $0.042 \times 0.050 = 2.1 \times 10^{-3}\ \text{mol}$

Adsorbed moles $= (3 - 2.1) \times 10^{-3} = 9 \times 10^{-4}\ \text{mol}$

Mass adsorbed (M of AcOH $= 60$) $= 9 \times 10^{-4} \times 60 = 0.054\ \text{g} = 54\ \text{mg}$

Per gram of charcoal $= \dfrac{54}{3} = \mathbf{18\ \text{mg/g}}$