Question

One gram of charcoal adsorbs 400 " mL of " 0.5 M acetic acid to form a mono layer and the molarity of acetic acid reduced to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. The surface area of charcoal is `3.01xx10^(2)m^(2)g^(-1)`.

Answer 1

Initial millomoles of `CH_(3)COOH=100xx0.5=50`
Final millimoles of `CH_(3)COOH=100xx0.49=49`
millimoles adsorbed `=50-49=1`
moles adsobed `=(1)/(1000)`
Molecules adsorbed`=(1)/(1000)xx6.023xx10^(23)`
`=6.023xx10^(20)`
Area per molecule`=("Total area")/("Number of molecules")`
`=(3.01xx10^(2))/(6.023xx10^(20))=5xx10^(19)m^(2)`