Verified Solution • 2025–26 Session moderate

QJEEAdv24Che7d4: JEE Advanced 2024 Chemistry - Class 12 JEEAdv24Che7d4 Solution: To form a complete monolayer of acetic acid on of charcoal, of…

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 (NA)=6.0×1023\left(\mathrm{N}_{\mathrm{A}}\right)=6.0 \times 10^{23} mol−1]\left.\mathrm{mol}^{-1}\right]

Step-by-Step Verified Solution

Let's first understand and break down the given information. We are provided with the following details to solve for the value of P\mathbf{P} :

1. Volume of acetic acid solution used: 100 mL100 \mathrm{~mL} of 0.5M0.5 \mathrm{M} acetic acid.

2. Volume of NaOH\mathrm{NaOH} solution used for neutralization: 40 mL40 \mathrm{~mL} of 1MNaOH1 \mathrm{M} \mathrm{NaOH} .

3. Surface area of the charcoal: 1.5×102 m2 g−11.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1} .

4. Avogadro's number: NA=6.0×1023 mol−1\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1} .

Let's start with the concentration and volume of acetic acid used:

Moles of acetic acid initially=0.5 M×100 mL1000=0.05 moles\text{Moles of acetic acid initially} = 0.5 \text{ M} \times \frac{100 \text{ mL}}{1000} = 0.05 \text{ moles}

Next, let's determine the moles of NaOH\mathrm{NaOH} used for neutralization:

Moles of NaOH=1 M×40 mL1000=0.04 moles\text{Moles of } \mathrm{NaOH} = 1 \text{ M} \times \frac{40 \text{ mL}}{1000} = 0.04 \text{ moles}

Since NaOH\mathrm{NaOH} completely neutralizes the unadsorbed acetic acid:

Moles of unadsorbed acetic acid=0.04 moles\text{Moles of unadsorbed acetic acid} = 0.04 \text{ moles}

Therefore, the moles of acetic acid adsorbed on the charcoal is given by:

Moles of adsorbed acetic acid=0.05−0.04=0.01 moles\text{Moles of adsorbed acetic acid} = 0.05 - 0.04 = 0.01 \text{ moles}

Now, we need to calculate the total number of molecules of acetic acid adsorbed:

Number of molecules=0.01 moles×6.0×1023 molecules/mole=6.0×1021 molecules\text{Number of molecules} = 0.01 \text{ moles} \times 6.0 \times 10^{23} \text{ molecules/mole} = 6.0 \times 10^{21} \text{ molecules}

Given the surface area of charcoal, we can find the total surface area:

Total surface area=1.5×102 m2/g×1 g=1.5×102 m2\text{Total surface area} = 1.5 \times 10^2 \text{ m}^2 / \text{g} \times 1 \text{ g} = 1.5 \times 10^2 \text{ m}^2

Finally, we need to determine the area occupied by each molecule of acetic acid:

Area per molecule of acetic acid=Total surface areaNumber of molecules=1.5×102 m26.0×1021 molecules\text{Area per molecule of acetic acid} = \frac{\text{Total surface area}}{\text{Number of molecules}} = \frac{1.5 \times 10^2 \text{ m}^2}{6.0 \times 10^{21} \text{ molecules}}

Let's perform the calculation:

Area per molecule of acetic acid=1.5×1026.0×1021=0.25×10−19 m2=2.5×10−20 m2\text{Area per molecule of acetic acid} = \frac{1.5 \times 10^2}{6.0 \times 10^{21}} = 0.25 \times 10^{-19} \text{ m}^2 = 2.5 \times 10^{-20} \text{ m}^2

Since the problem states that the area per molecule is P×10−23 m2\mathbf{P} \times 10^{-23} \mathrm{~m}^2 , we equate:

2.5×10−20=P×10−232.5 \times 10^{-20} = \mathbf{P} \times 10^{-23}

Solving for P\mathbf{P} :

P=2.5×10−2010−23=2.5×103=2500\mathbf{P} = \frac{2.5 \times 10^{-20}}{10^{-23}} = 2.5 \times 10^3 = 2500

Therefore, the value of P\mathbf{P} is 2500.

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