To form a complete monolayer of acetic acid on 1g 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 × 10^-23 m²  surface area on charcoal, the value of P is _____.

$[\text{Use given data : Surface area of charcoal}=1.5\times {10}^2\text{m}{2}^{g-1},$

$\text{Avogadro’s number}\left(N_A\right)=6.0\times {10}^{23}\text{mol}{\mathrm{^-1}}^{}]$

The correct answer is 2500.

Let us solve the problem step-by-step:

Step 1: Calculate the initial millimoles and moles of acetic acid.
Volume of acetic acid solution = 100 mL
Molarity of acetic acid solution = 0.5 M

$\text{Initial moles of acetic acid}=\text{Molarity}\times \text{Volume (in L)}=0.5\times \frac{100}{1000}=0.05\text{mol}$

Step 2: Calculate the unadsorbed moles of acetic acid.
To neutralize the unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required.
Since 1 mole of acetic acid reacts with 1 mole of NaOH (CH₃COOH + NaOH ⇒ CH₃COONa + H₂O):

$\text{Unadsorbed moles of acetic acid}=\text{Moles of NaOH used}=1.0\times \frac{40}{1000}=0.04\text{mol}$

Step 3: Calculate the moles and number of molecules of acetic acid adsorbed.

$\text{Moles of acetic acid adsorbed}=\text{Initial moles}-\text{Unadsorbed moles}=0.05-0.04=0.01\text{mol}$

Using Avogadro's number (N_A = 6.0 × 10²³ mol^-1):

$\text{Number of adsorbed molecules}=0.01\times 6.0\times {10}^{23}=6.0\times {10}^{21}\text{molecules}$

Step 4: Calculate the surface area occupied by each molecule.
Surface area of 1 g of charcoal = 1.5 × 10² m² g^-1 = 150 m²
Since a complete monolayer is formed on 1 g of charcoal, the total area covered by all adsorbed molecules is equal to the total surface area of charcoal.

$\text{Area occupied per molecule}=\frac{\text{Total surface area of charcoal}}{\text{Total number of adsorbed molecules}}$

$\text{Area per molecule}=\frac{150}{6.0\times {10}^{21}}=25\times {10}^{-21}{\text{m}}^2=2500\times {10}^{-23}{\text{m}}^2$

Given that the surface area occupied by each molecule is P × 10^-23 m², comparing both values gives:
P = 2500