A quantity of 1.0 g of charcoal adsorbs 100 ml 0.5 M CH₃COOH to form a monolayer, and thereby the molarity of CH₃COOH reduces to 0.49 M. The surface area (in 10⁻¹⁹ m²) of the charcoal adsorbed by each

Question

A quantity of 1.0 g of charcoal adsorbs 100 ml 0.5 M CH₃COOH to form a monolayer, and thereby the molarity of CH₃COOH reduces to 0.49 M. The surface area (in 10⁻¹⁹ m²) of the charcoal adsorbed by each molecule of acetic acid, is (Surface area of charcoal = 3.01 × 10² m²/g).

Prashant Kotian · Accepted Answer

Correct answer: 5. Initial moles of CH₃COOH = 0.1 L × 0.5 M = 0.05 mol. Final moles = 0.1 L × 0.49 M = 0.049 mol. Moles of acetic acid adsorbed = 0.05 - 0.049 = 0.001 mol. Number of molecules adsorbed = 0.001 × 6.02 × 10²³ = 6.02 × 10²⁰ molecules. Total surface area of 1 g charcoal = 3.01 × 10² m². Surface area occupied per molecule = (3.01 × 10² m²) / (6.02 × 10²⁰ molecules) = 0.5 × 10⁻¹⁸ m² = 5 × 10⁻¹⁹ m². The value in terms of 10⁻¹⁹ m² is 5. Therefore, correct answer is 5.

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