To form a complete monolayer of acetic acid on $1 \text{ g}$ of charcoal, $100 \text{ mL}$ of $0.5 \text{ M}$ acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, $40 \text{ mL}$ of $1 \text{ M NaOH}$ solution was required. If each molecule of ...

The correct numerical answer is 2500. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Chemistry Question 8 Solution

Initial moles of acetic acid $= \text{Molarity} \times \text{Volume (L)} = 0.5 \times 0.1 = 0.05 \text{ mol}$ .
Moles of unadsorbed acetic acid (neutralized by NaOH) $= 1 \times 0.04 = 0.04 \text{ mol}$ .
Moles of acetic acid adsorbed on charcoal $= 0.05 - 0.04 = 0.01 \text{ mol}$ .
Number of molecules adsorbed $= \text{moles} \times N_A = 0.01 \times 6.0 \times 10^{23} = 6.0 \times 10^{21} \text{ molecules}$ .
Total surface area of $1 \text{ g}$ charcoal $= 1.5 \times 10^2 \text{ m}^2 = 150 \text{ m}^2$ .
Surface area per molecule $= \frac{\text{Total Area}}{\text{Number of molecules}} = \frac{150}{6.0 \times 10^{21}} = 25 \times 10^{-21} \text{ m}^2$ .
Converting to the required form: $25 \times 10^{-21} = 2500 \times 10^{-23} \text{ m}^2$ .
Therefore, $P = 2500$ .