Chemistry · Surface Chemistry

JEE Advanced 2024 — Paper 2 — Question 30

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 $\mathbf{P} \times 10^{-23} \mathrm{~m}^{2}$ surface area on charcoal, the value of $\mathbf{P}$ is $\_\_\_\_\_$ _. [Use given data : Surface area of charcoal $=1.5 \times 10^{2} \mathrm{~m}^{2} \mathrm{~g}^{-1}$ ; Avogadro's number $\left(N_{A}\right)=6.0 \times 10^{23} \mathrm{~mol}^{-1}$ ]

Answer: 2500

Numerical answer — enter this value.

Step-by-step solution

Number of moles of $\mathrm{NaOH}=40 \times 1 \times 10^{-3}=4 \times 10^{-2}$ Number of moles of Acetic acid $=100 \times 0.5 \times 10^{-3}=5 \times 10^{-2}$ $\therefore$ Number of moles of acetic acid adsorbed on charcoal $=5 \times 10^{-2}-4 \times 10^{-2}=1 \times 10^{-2}$ $\therefore$ Number of molecules of acetic acid adsorbed $=1 \times 10^{-2} \times \mathrm{N}_{\mathrm{A}}$

=6 \times 10^{21}

Total surface area adsorbed by one molecule of acetic acid $=\frac{1.5 \times 10^{2}}{6 \times 10^{21}}=2500 \times 10^{-23}$ $\therefore \mathrm{P}=2500$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Chemistry
Chapter: Surface Chemistry
Topic: Adsorption