Question

When 200 mL of 0.2 M acetic acid is shaken with 0.6 g of wood charcoal, the final concentration of acetic acid after adsorption is 0.1 M. The mass of acetic acid adsorbed per gram of carbon is _____ g.

Answer 1

Correct Answer: 2

To solve the problem, we need to determine the mass of acetic acid adsorbed per gram of carbon.

1. Determine initial moles of acetic acid: Use the formula \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). Given the initial volume is 200 mL (0.2 L) and the molarity is 0.2 M, we calculate:

\( \text{moles}_{\text{initial}} = 0.2 \times 0.2 = 0.04 \, \text{mol} \).

2. Determine final moles of acetic acid: After adsorption, the concentration is 0.1 M, so the moles are:

\( \text{moles}_{\text{final}} = 0.2 \times 0.1 = 0.02 \, \text{mol} \).

3. Calculate moles of acetic acid adsorbed: Subtract the final moles from the initial moles:

\( \text{moles}_{\text{adsorbed}} = 0.04 - 0.02 = 0.02 \, \text{mol} \).

4. Convert moles to grams: The molar mass of acetic acid (CH₃COOH) is approximately 60 g/mol. Thus, the mass adsorbed is:

\( \text{mass}_{\text{adsorbed}} = 0.02 \times 60 = 1.2 \, \text{g} \).

5. Determine adsorbed mass per gram of carbon: Given 0.6 g of charcoal is used, compute the adsorbed mass per gram:

\( \text{mass}_{\text{per g}} = \frac{1.2}{0.6} = 2.0 \, \text{g/g} \).

6. Verify within range: The computed value of 2.0 g/g fits within the given range of 2–2.

Conclusion: The mass of acetic acid adsorbed per gram of carbon is 2.0 g.