Area Occupied by Acetic Acid Monolayer on Charcoal Surface

To determine the surface area occupied by each molecule of acetic acid on charcoal, we calculate the number of moles of acetic acid adsorbed onto the charcoal surface as follows:

Step 1: Calculate initial moles of acetic acid $\text{Initial moles of } \text{CH}_3\text{COOH} = M_1 \times V_1 = 0.5 \text{ M} \times 100 \times 10^{-3} \text{ L} = 0.05 \text{ mol}$

Step 2: Calculate unadsorbed moles of acetic acid The neutralization reaction between acetic acid and sodium hydroxide is: $\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$

Thus, 1 mole of $\text{NaOH}$ neutralizes 1 mole of unadsorbed $\text{CH}_3\text{COOH}$. $\text{Unadsorbed moles of } \text{CH}_3\text{COOH} = \text{Moles of } \text{NaOH} = M_2 \times V_2 = 1 \text{ M} \times 40 \times 10^{-3} \text{ L} = 0.04 \text{ mol}$

Step 3: Calculate moles and molecules of acetic acid adsorbed $\text{Moles of } \text{CH}_3\text{COOH} \text{ adsorbed} = 0.05 \text{ mol} - 0.04 \text{ mol} = 0.01 \text{ mol}$

Using Avogadro's number ($N_{\text{A}} = 6.0 \times 10^{23} \text{ mol}^{-1}$), the number of adsorbed molecules ($N$) is: $N = 0.01 \text{ mol} \times 6.0 \times 10^{23} \text{ mol}^{-1} = 6.0 \times 10^{21} \text{ molecules}$

Step 4: Calculate the area occupied per molecule The total surface area of $1\text{ g}$ of charcoal is given as: $A_{\text{total}} = 1.5 \times 10^2 \text{ m}^2 = 150 \text{ m}^2$

The area occupied by a single molecule ($a$) is: $a = \frac{A_{\text{total}}}{N} = \frac{150 \text{ m}^2}{6.0 \times 10^{21}} = 2.5 \times 10^{-20} \text{ m}^2$

Expressing this in the form $\mathbf{P} \times 10^{-23} \text{ m}^2$: $a = 2500 \times 10^{-23} \text{ m}^2$

Thus, the value of $\mathbf{P}$ is 2500.