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

To solve the problem, we need to determine the value of \(P\), where the area occupied by each acetic acid molecule adsorbed on charcoal is \(P \times 10^{-23} \, \text{m}^2\).

1. Calculate the Moles of Acetic Acid Initially Present:
The initial acetic acid solution has:
- Volume = 100 mL = 0.1 L
- Molarity = 0.5 M
Moles of acetic acid:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} = 0.5 \, \text{M} \times 0.1 \, \text{L} = 0.05 \, \text{moles} \]

2. Calculate the Moles of Unadsorbed Acetic Acid:
The unadsorbed acetic acid is neutralized by NaOH:
- Volume of NaOH = 40 mL = 0.04 L
- Molarity of NaOH = 1 M
Moles of NaOH:
\[ \text{Moles of NaOH} = 1 \, \text{M} \times 0.04 \, \text{L} = 0.04 \, \text{moles} \] Since NaOH reacts with acetic acid in a 1:1 molar ratio, moles of unadsorbed acetic acid = 0.04 moles.

3. Calculate the Moles of Acetic Acid Adsorbed:
\[ \text{Moles adsorbed} = \text{Initial moles} - \text{Unadsorbed moles} = 0.05 - 0.04 = 0.01 \, \text{moles} \]

4. Calculate the Number of Acetic Acid Molecules Adsorbed:
Using Avogadro’s number (\(6.0 \times 10^{23} \, \text{molecules/mole}\)):
\[ \text{Number of molecules} = 0.01 \, \text{moles} \times 6.0 \times 10^{23} \, \text{molecules/mole} = 6.0 \times 10^{21} \, \text{molecules} \]

5. Calculate the Total Surface Area of the Charcoal:
Given:
- Surface area of charcoal = \(1.5 \times 10^2 \, \text{m}^2/\text{g}\)
- Mass of charcoal = 1 g
\[ \text{Total surface area} = 1.5 \times 10^2 \, \text{m}^2/\text{g} \times 1 \, \text{g} = 1.5 \times 10^2 \, \text{m}^2 \]

6. Calculate the Area Occupied by One Acetic Acid Molecule:
\[ \text{Area per molecule} = \frac{\text{Total surface area}}{\text{Number of molecules}} = \frac{1.5 \times 10^2 \, \text{m}^2}{6.0 \times 10^{21} \, \text{molecules}} = 2.5 \times 10^{-20} \, \text{m}^2/\text{molecule} \]

7. Determine the Value of \(P\):
The area per molecule is given as \(P \times 10^{-23} \, \text{m}^2\). Thus:
\[ P \times 10^{-23} = 2.5 \times 10^{-20} \] \[ P = \frac{2.5 \times 10^{-20}}{10^{-23}} = 2.5 \times 10^3 = 2500 \]

Final Answer:
The value of \(P\) is \(2500\).

Answer 2

To solve the problem, we need to find the value of \( P \) where each molecule of acetic acid occupies \( P \times 10^{-23} \, \text{m}^2 \) on charcoal.

1. Calculate the initial moles of acetic acid added:
Volume of acetic acid solution = 100 mL = 0.1 L
Concentration of acetic acid = 0.5 M = 0.5 mol/L
Initial moles of acetic acid:
\[ n_{\text{initial}} = 0.5 \times 0.1 = 0.05 \, \text{mol} \]

2. Calculate moles of unadsorbed acetic acid:
Volume of NaOH used = 40 mL = 0.04 L
Concentration of NaOH = 1 M = 1 mol/L
Moles of NaOH = moles of unadsorbed acetic acid (neutralization):
\[ n_{\text{unadsorbed}} = 1 \times 0.04 = 0.04 \, \text{mol} \]

3. Calculate moles of acetic acid adsorbed:
\[ n_{\text{adsorbed}} = n_{\text{initial}} - n_{\text{unadsorbed}} = 0.05 - 0.04 = 0.01 \, \text{mol} \]

4. Calculate total number of adsorbed molecules:
\[ N = n_{\text{adsorbed}} \times N_A = 0.01 \times 6.0 \times 10^{23} = 6.0 \times 10^{21} \, \text{molecules} \]

5. Surface area covered by adsorbed molecules equals surface area of charcoal:
Surface area of charcoal = \(1.5 \times 10^{2} \, \text{m}^2 \)

6. Calculate surface area occupied per molecule:
\[ \text{Area per molecule} = \frac{\text{Total surface area}}{\text{Number of molecules}} = \frac{1.5 \times 10^{2}}{6.0 \times 10^{21}} = 2.5 \times 10^{-20} \, \text{m}^2 \]

7. Express area per molecule in the form \(P \times 10^{-23} \, \text{m}^2\):
\[ 2.5 \times 10^{-20} = P \times 10^{-23} \Rightarrow P = \frac{2.5 \times 10^{-20}}{10^{-23}} = 2.5 \times 10^{3} = 2500 \]

Final Answer:
The value of \( P \) is \(\boxed{2500}\).