1g of charcoal adsorbs 100 mL 0.5 M CH₃COOH to form a monolayer, and thereby the molarity of CH₃COOH reduces to 0.49. Calculate the surface area of the charcoal covered by each molecule of acetic acid. Surface area of charcoal $= 3 .01 \times 10^{2} m^{2} / g$

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$4 \times 10^{- 20} m^{2}$

$0.5 \times 10^{- 19} m^{2}$

$5 \times 10^{- 19} m^{2}$

$0.5 \times 10^{- 10} m^{2}$

answer is D.

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Detailed Solution

Amount of CH₃COOH adsorbed on lg of charcoal $= (0 .5 - 0 .49 {molL}^{- 1}) \times \frac{1000 mL}{1000 mL / L} = 1 \times 10^{- 3} mol$
So, No. of CH₃COOH molecules in 1 x 10^-3 mol of acetic acid $= 6 .02 \times 10^{23} {mol}^{- 1} \times 10^{- 3} mol = 6 .02 \times 10^{20}$
Area occupied by one CH₃COOH molecule = $\frac{3 .01 \times 10^{2} m^{2}}{6 .02 \times 10^{20}} = 5 \times 10^{- 19} m^{2}$
Thus the surface area of the charcoal covered by each CH₃COOH molecule is 5 x 10^-19 m².