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 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 ]

The correct answer is 2500.

Let us find the value of $P$ step-by-step by determining the amount of acetic acid adsorbed on the charcoal surface.

Step 1: Calculate the initial moles of acetic acid
The volume of the acetic acid solution used is 100 mL, and its molarity is 0.5 M.
Initial moles of acetic acid = Molarity $\times$ Volume in liters
$\text{Initial moles}=0.5\text{mol/L}\times \frac{100}{1000}\text{L}=0.05\text{mol}$

Step 2: Calculate the moles of unadsorbed acetic acid
The unadsorbed acetic acid is neutralized by 40 mL of 1 M NaOH solution. The neutralization reaction between acetic acid (${\text{CH}}_3\text{COOH}$) and NaOH is 1:1.
Therefore, the moles of unadsorbed acetic acid equal the moles of NaOH used.
$\text{Moles of unadsorbed acetic acid}=\text{Moles of NaOH}=1\text{mol/L}\times \frac{40}{1000}\text{L}=0.04\text{mol}$

Step 3: Calculate the moles of acetic acid adsorbed on charcoal
The moles of acetic acid adsorbed to form a monolayer is the difference between the initial moles and the unadsorbed moles.
$\text{Moles of adsorbed acetic acid}=0.05\text{mol}-0.04\text{mol}=0.01\text{mol}$

Step 4: Calculate the total number of adsorbed molecules
Using Avogadro's number ($N_A=6.0\times {10}^{23}{\text{mol}}^{-1}$):
$\text{Number of adsorbed molecules}=0.01\text{mol}\times 6.0\times {10}^{23}{\text{mol}}^{-1}=6.0\times {10}^{21}\text{molecules}$

Step 5: Relate the number of molecules to the surface area
The total surface area of 1 g of charcoal is given as $1.5\times {10}^2{\text{m}}^2{\text{g}}^{-1}$.
Since a complete monolayer is formed on 1 g of charcoal, the total area occupied by the adsorbed molecules is equal to the surface area of the charcoal.
Let $A$ be the surface area occupied by a single molecule of acetic acid:
$A=P\times {10}^{-23}{\text{m}}^2$
Therefore, the total area is:
$\text{Total area}=\left(\text{Number of molecules}\right)\times A$
$1.5\times {10}^2{\text{m}}^2=(6.0\times {10}^{21})\times (P\times {10}^{-23}{\text{m}}^2)$

Step 6: Solve for P
Simplify the equation:
$1.5\times {10}^2=6.0\times P\times {10}^{21-23}$
$150=6.0\times P\times {10}^{-2}$
$150=0.06\times P$
$P=\frac{150}{0.06}=2500$

Thus, the value of $P$ is 2500.