1 gmof charcoal adsorbs 100 mL 0.7 M CH_(3)COOHto form a monolayer, and there by the molarity of CH_(3)COOHreduces to 0.59. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid in terms of 10^(-19). Surface area of charcoal = 3.01 xx 10^(2) m^(2)//gm.

1 gmof charcoal adsorbs 100 mL 0.7 M CH_(3)COOHto form a monolayer, an

1.	1 gmof charcoal adsorbs 100 mL 0.7 M CH_(3)COOHto form a monolayer, and there by the molarity of CH_(3)COOHreduces to 0.59. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid in terms of 10^(-19). Surface area of charcoal = 3.01 xx 10^(2) m^(2)//gm.
Answer» Solution : Number of moles of ACETIC acid in 100 mL before adding charcoal = 0.07 Number of moles of acetic acid in 100 ml after adding charcoal = 0.059 Number of moles of acetic acid adsorbed on the surface of charcoal. Number of MOLECULES of acetic acid adsorbed on the surface ofcharcoal = `0.011xx 6.02 XX 10^23 =6.62 xx 10^21` Surface area of charcoal = `3.01 xx 10^2 m^2` (given) Area occupied by single acetic acid molecule on the surface of charcoal `=(3.01 xx 10^(2))/(6.62 xx 10^(21)) = 0.45 xx 10^(-19) m^(2)`