To form a complete monolayer of acetic acid on 1 g of activated charcoal, 100 mL of 0.5 M acetic acid solution was used. Some acetic acid remained unadsorbed. To neutralise this unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies P * 10

Question

To form a complete monolayer of acetic acid on 1 g of activated charcoal, 100 mL of 0.5 M acetic acid solution was used. Some acetic acid remained unadsorbed. To neutralise this unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies P * 10⁻²¹ m² of surface area on the charcoal, find the value of P/5. [Surface area of charcoal = 1.5 * 10² m² per gram; Avogadro number = 6.0 * 10²³ mol⁻¹]

Accepted Answer

5. Initial moles of acetic acid = 0.1 L * 0.5 mol/L = 0.05 mol. Moles unadsorbed = 0.04 L * 1 mol/L = 0.04 mol. Moles adsorbed = 0.05 - 0.04 = 0.01 mol. Molecules adsorbed = 0.01 * 6.0*10²³ = 6.0*10²¹ molecules. Surface area of 1 g charcoal = 1.5*10² m² = 150 m². Area per molecule = 150 / (6.0*10²¹) = 25*10⁻²¹ m². So P = 25, and P/5 = 5.

Solution

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