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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Advanced 2024 · Paper 2 · Chemistry · Some Basic Concepts of Chemistry

To form a complete monolayer of acetic acid on 1 ~g of charcoal,...

To form a complete monolayer of acetic acid on 1 g1 \mathrm{~g} of charcoal, 100 mL100 \mathrm{~mL} of 0.5M0.5 \mathrm{M} acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL\mathrm{mL} of 1MNaOH1 \mathrm{M} \mathrm{NaOH} solution was required. If each molecule of acetic acid occupies P×10−23 m2\mathbf{P} \times 10^{-23} \mathrm{~m}^2 surface area on charcoal, the value of P\mathbf{P} is _______ [Use given data: Surface area of charcoal =1.5×102 m2 g−1;=1.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1} ; Avogadro's number \left(\mathrm{N}_{\mathrm{A}}\right)=6.0 \times 10^{23}mol−1]\left.\mathrm{mol}^{-1}\right]

Correct answer

0

Step-by-step solution

 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 =1.5×1026×1021=0.25×10−19=2500×10−23\begin{aligned} & \text{ Millimole of acid taken }=100 \times 0.5=50 \\ & \text{ Millimole of } \mathrm{NaOH} \text{ used }=40 \times 1=40 \\ & \text{ Millimole of acid adsorbed }=50-40=10 \\ & \text{ Molecules of acid adsorbed }=10 \times 10^{-3} \times 6 \times 10^{23}=6 \times 10^{21} \\ & \text{ Surface area occupied per molecule }=\frac{1.5 \times 10^2}{6 \times 10^{21}}=0.25 \times 10^{-19}=2500 \times 10^{-23}\end{aligned}
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