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JEE Advanced2024ChemistrySome Basic Concepts of ChemistryActual

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 10⁻²³ ~m ^2 surface area on charcoal, the value of P is _______ [Use given data: Surface area of charcoal =1.5 10^2 ~m ^2 ~g ⁻¹ ; Avogadro's numb

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Step-by-step solution

aligned & Millimole of acid taken =100 0.5=50 & Millimole of NaOH used =40 1=40 & Millimole of acid adsorbed =50-40=10 & Molecules of acid adsorbed =10 10⁻³ 6 10²³=6 10²¹ & Surface area occupied per molecule = 1.5 10^2 6 10²¹ =0.25 10⁻¹⁹=2500 10⁻²³ aligned

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