JEE Advanced 2024 — Paper 2 — CHEM-III solution

JEE Advanced2024Paper 2CHEM-III

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 $N a O H$ solution was required. If each molecule of acetic acid occupies $P \times 1 0^{- 23}$ m $^{2}$ surface area on charcoal, the value of $P$ is________.

[Use given data: Surface area of charcoal $= 1.5 \times 1 0^{2}$ m $^{2}$ g $^{- 1}$ ; Avogadro's number ( $N_{A}$ ) $= 6.0 \times 1 0^{23}$ mol $^{- 1}$ ]

Your answer

Solution

Initial moles of acetic acid taken: $0.1 L \times 0.5 M = 0.05$ mol.

Moles of unadsorbed acetic acid (titrated by NaOH): $0.040 L \times 1 M = 0.04$ mol.

Moles of acetic acid adsorbed: $0.05 - 0.04 = 0.01$ mol.

Number of acetic acid molecules adsorbed: $0.01 \times 6.0 \times 1 0^{23} = 6.0 \times 1 0^{21} .$

Total surface area of $1$ g of charcoal: $1.5 \times 1 0^{2} m^{2} .$

Area per molecule:

$a = \frac{1.5 \times 1 0 ^{2}}{6.0 \times 1 0 ^{21}} = 2.5 \times 1 0^{- 20} m^{2} = 2500 \times 1 0^{- 23} m^{2} .$

$∴ P = 2500.$

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