1 g of charcoal adsorbs 100 ml 0.5 M $CH_3COOH$ to form a monolayer, and thereby the molarity of $CH

Question

1 g of charcoal adsorbs 100 ml 0.5 M $CH_3COOH$ to form a monolayer, and thereby the molarity of $CH_3COOH$ reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = $3.0 	imes 10^2~m^2/g$.

Professor PC Thomas Classes · Answer

Moles adsorbed = $(0.5 - 0.49) 	imes 0.1 = 0.001$ mol. Molecules = $0.001 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{20}$. Area/molecule = $300 / (6.022 	imes 10^{20}) \approx 5 	imes 10^{-19}~m^2$.