One gram of charcoal adsorbs 100ml of `0.5M CH_(3)COOH` to form a monolayer and thereby the molarity of acidic acid is reduced to 0.49M. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid (surface area of charcoal is `3.01xx10^(2) m^(2)//g`).

Explanation

Number of moles of acetic acid before the adsorption `=0.5xx(100)/(1000)=0.5`
Number of moles of acetic acid after the adsorption `=0.49xx(100)/(1000)=0.049`
  Number of moles of acetic acid adsorbed `= 0.05-0.049 = 0.001`.
Number of molecules of acetic acid adsorbed` = 0.001 xx6.023xx 10^(2) = 6.023 x 10^(20)`
Surface area of the charcoal occupied by each acetic acid molecule `=(3.01xx10^(2))/(6.023xx10^(20))=5xx10^(-19)m^(2)`.