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JEE Advanced 2024 · Paper 2, Chemistry question 8

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 P×10−23 m2\mathbf{P}\times10^{-23}\,\mathrm{m^2} surface area on charcoal, the value of P is _____.

[Use given data: Surface area of charcoal =1.5×102 m2 g−1= 1.5\times10^2\,\mathrm{m^2\,g^{-1}}; Avogadro's number (NA)=6.0×1023 mol−1(N_A) = 6.0\times10^{23}\,\mathrm{mol^{-1}}]

Answer and worked solution

Official answer: 2500

Moles adsorbed.

  • Acetic acid taken: 0.100×0.5=0.0500.100\times0.5=0.050 mol.
  • Left unadsorbed (equal to the NaOH used): 0.040×1=0.0400.040\times1=0.040 mol.
  • Adsorbed: 0.0100.010 mol, i.e. 0.010×6.0×1023=6.0×10210.010\times6.0\times10^{23}=6.0\times10^{21} molecules.

Area per molecule. The monolayer covers 1.5×102 m21.5\times10^2\ \mathrm{m^2}, so
1506.0×1021=2.5×10−20 m2=2500×10−23 m2.\frac{150}{6.0\times10^{21}}=2.5\times10^{-20}\ \mathrm{m^2}=2500\times10^{-23}\ \mathrm{m^2}.

So P=2500\mathbf P=2500.

Topics:AdsorptionMole conceptIndicators and titrations
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