To form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5 M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies $$P \times 10^{-23}$$ $$m^2$$ surface area on charcoal, the value of P is ________.

[Use given data : Surface area of charcoal = $$1.5 \times 10^2$$ $$m^2 g^{-1}$$; Avogadro's number ($$N_A$$) = $$6.0 \times 10^{23}$$ $$mol^{-1}$$]

Total surface area available for adsorption on $$1 \text{ g}$$ of charcoal is given to be
$$A = 1.5 \times 10^{2}\; \text{m}^{2}$$

Step 1: Moles of acetic acid taken
Solution volume = $$100\;\text{mL} = 0.1\;\text{L}$$
Molarity = $$0.5\;\text{M}$$
$$n_{\text{taken}} = 0.1 \times 0.5 = 0.05\;\text{mol}$$

Step 2: Moles of acetic acid remaining unadsorbed
Volume of NaOH used = $$40\;\text{mL} = 0.04\;\text{L}$$
Molarity of NaOH = $$1\;\text{M}$$
For $$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$ the stoichiometry is 1 : 1, so
$$n_{\text{unadsorbed}} = 0.04 \times 1 = 0.04\;\text{mol}$$

Step 3: Moles of acetic acid adsorbed on charcoal
$$n_{\text{adsorbed}} = n_{\text{taken}} - n_{\text{unadsorbed}} = 0.05 - 0.04 = 0.01\;\text{mol}$$

Step 4: Number of molecules adsorbed
Avogadro number $$N_A = 6.0 \times 10^{23}\;\text{mol}^{-1}$$
$$N = 0.01 \times 6.0 \times 10^{23} = 6.0 \times 10^{21}$$ molecules

Step 5: Surface area occupied by one molecule
$$\text{Area per molecule} = \frac{A}{N} = \frac{1.5 \times 10^{2}}{6.0 \times 10^{21}} = 0.25 \times 10^{-19}\;\text{m}^{2} = 2.5 \times 10^{-20}\;\text{m}^{2}$$

Step 6: Expressing in the required form
$$2.5 \times 10^{-20}\;\text{m}^{2} = 2.5 \times 10^{3} \times 10^{-23}\;\text{m}^{2}$$

Hence $$P = 2500$$.

Final answer: 2500