Question

One gram of charcoal adsorbs 100 mL of 0.5 `M CH_(3)COOH` to form a mono-layer and thereby the molarity of acetic acid is reduced to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface acid of charcoal `=3.01xx10^(2) m^(2)//gm`

Answer 1

Number of moles of acetic acid initially present
`(MV)/(100)=(0.5xx100)/(1000) =0.05`
Number of moles of acetic acid left
`=(MV)/(1000)=(0.5xx100)/(1000)=0.049`
Number of moles of acetic acid adsorbed
=0.05-0.049=0.001 mol
Number of molecules of acid adsorbed
`=0.001xx6.023xx10^(23)=6.023xx10^(20)`
Area occupied by single molecule of acetic acid
`=("Total area")/("Number of molecules adsorbed")`
`=(3.01xx10^(2))/(6.023xx10^(20))`
`=5xx10^(-19)m^(2)`