JEE Advanced · Chemistry · 1. Mole Concept
To form a complete monolayer of acetic acid on \(1 \mathrm{~g}\) of charcoal, \(100 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 \(\mathrm{mL}\) of \(1 \mathrm{M} \mathrm{NaOH}\) 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 \(\left(\mathrm{N}_{\mathrm{A}}\right)=6.0 \times 10^{23}\) \(\left.\mathrm{mol}^{-1}\right]\)
- A 2000
- B 250
- C 3241
- D 2500
Answer & Solution
Correct Answer
(D) 2500
Step-by-step Solution
Detailed explanation
\(\text {Millimole of acid taken }=100 \times 0.5=50 \)
\( \text {Millimole of } \mathrm{NaOH} \text { used }=40 \times 1=40 \)
\( \text {Millimole of acid adsorbed }=50-40=10 \)
\( \text {Molecules of acid adsorbed }=10 \times 10^{-3} \times 6\) \(\times~10^{23}=6 \times 10^{21} \)
\( \text {Surface area occupied per molecule }\) \(=\frac{1.5 \times 10^2}{6 \times 10^{21}}=0.25 \times 10^{-19}=2500 \times 10^{-23}\)
\( \text {Millimole of } \mathrm{NaOH} \text { used }=40 \times 1=40 \)
\( \text {Millimole of acid adsorbed }=50-40=10 \)
\( \text {Molecules of acid adsorbed }=10 \times 10^{-3} \times 6\) \(\times~10^{23}=6 \times 10^{21} \)
\( \text {Surface area occupied per molecule }\) \(=\frac{1.5 \times 10^2}{6 \times 10^{21}}=0.25 \times 10^{-19}=2500 \times 10^{-23}\)
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