Question

To form a complete monolayer of acetic acid on 1g of charcoal, 100 mL of 0.5M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1M NaOH solution was required. If each molecule of acetic acid occupies P x 10⁻²³ m² surface area on charcoal, the value of P is _____.
[Use given data: Surface area of charcoal = 1.5 x 10² m²g⁻¹ ; Avogadro’s number (N_A) = 6.0 x 10²³ mol⁻¹ ]

Answer 1

Correct Answer: 2500

Calculation of Moles and Surface Area Occupied by Adsorbed Molecules 

1. Moles of Acetic Acid Initially Present:

\[ \text{Concentration} \times \text{Volume} = 0.5M \times 100 \, \text{mL} = 0.05 \, \text{mol} \]

2. Moles of Unadsorbed Acetic Acid Neutralized by NaOH:

\[ \text{Concentration} \times \text{Volume} = 1M \times 40 \, \text{mL} = 0.04 \, \text{mol} \]

3. Moles of Acetic Acid Adsorbed:

\[ 0.05 \, \text{mol} - 0.04 \, \text{mol} = 0.01 \, \text{mol} \]

4. Number of Molecules Adsorbed:

\[ 0.01 \, \text{mol} \times 6.022 \times 10^{23} = 6.022 \times 10^{21} \, \text{molecules} \]

5. Surface Area Occupied by One Molecule:

\[ P \times 10^{-23} = \frac{\text{Surface Area of Charcoal}}{\text{Number of Molecules Adsorbed}} \]

Substituting values:

\[ P \times 10^{-23} = \frac{1.5 \times 10^2}{6.022 \times 10^{21}} \]

Quick Tip:

\[ P = 2500 \]

Final Answer:

The surface area occupied by one molecule is 2500 units.