Surface Chemistry · Question 3

JEE Advanced 2024 Paper 2NumericalMedium📈 Topic in 17% papers+4

To form a complete monolayer of acetic acid on 1g 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 m2 surface area on charcoal, the value of P is . [Use given data: Surface area of charcoal = 1.5 102 m2 g-1; Avogadro’s number (NA) = 6.0 1023 mol-1]

Answer

2500

Solution

Moles of acetic acid taken = 0.1 0.5 = 0.05 mol

Unadsorbed = moles of NaOH = 0.040 1 = 0.04 mol

Adsorbed = 0.01 mol = 0.01 6.0 1023 = 6.0 1021 molecules

Area per molecule = 1.5 1026.0 1021 = 2.5 10-20 m2 = 2500 10-23 m2

Answer: 2500

JEE Advanced 2024 Paper 2

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At a given temperature, 0.45 g of acetic acid in 50 mL of water is shaken with 1.0 g of charcoal and the pH of the resulting solution is 3.0. Assume, the adsorption of acetic acid from the aqueous solution by charcoal follows Freundlich isotherm, xm=kC1/n If the plot of 10(x/m) against 10C gives a straight line with slope 1, the value of k in L mol-1 is . Given: The molar mass of acetic acid is 60 g mol-1. The acid dissociation constant of acetic acid is 1.0 10-5 at the given temperature. x is the mass (in grams) of acetic acid adsorbed. m is the mass (in grams) of charcoal. C is the equilibrium concentration of acetic acid in the solution after the adsorption is complete. k and n are constants for acetic acid–charcoal system at the given temperature.

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