Question

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

Hint:

To calculate adsorption, determine the moles of the substance adsorbed and use Avogadro’s number to calculate the number of molecules. Divide the total surface area by the number of molecules to find the surface area occupied per molecule.

Answer 1

1. Moles of acetic acid initially present: \[ \text{Concentration} \times \text{Volume} = 0.5 \, \text{M} \times 100 \, \text{mL} = 0.05 \, \text{mol}. \] 2. Moles of unadsorbed acetic acid neutralized by NaOH: \[ \text{Concentration} \times \text{Volume} = 1 \, \text{M} \times 40 \, \text{mL} = 0.04 \, \text{mol}. \] 3. Moles of acetic acid adsorbed: \[ 0.05 \, \text{mol} - 0.04 \, \text{mol} = 0.01 \, \text{mol}. \] 4. Number of molecules adsorbed: \[ 0.01 \, \text{mol} \times 6.022 \times 10^{23} = 6.022 \times 10^{21} \, \text{molecules}. \] 5. Surface area occupied by one molecule: \[ P \times 10^{-23} = \frac{\text{Surface area of charcoal}}{\text{Number of molecules adsorbed}}. \] Substituting values: \[ P \times 10^{-23} = \frac{1.5 \times 10^2}{6.022 \times 10^{21}}. \] \[ P = 2500. \]