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`

Explanation

No of m mole of `CH_(3)COOH=100xx05=50`
Since concentriation reduces to `0.49`M
`:.` Final no. of m mole of `CH_(3)COOH=100xx0.49=49`
`:.` no of m molecules of `CH_(3)COOH` get adsorbed `=50-49=1`
`:.` No of molecules of `CH_(3)COOH` get adsorbed
`=6.02xx10^(23)xx10^(-3)=6.02xx10^(20)`
since 1g characoal has area `=3.01xx10^(2)m^(2)`
`:. 6.02 xx10^(20)` molecules of acetic acid gets adsorned in` 3.01xx10^(2) m^(2)` area
`:.` 1 molecules of acetic acid gets adsorbe `=(3.01xx10^(2))/(6.02xx10^(20))`
`=(1)/(2)xx10^(-18)=5xx10^(-19)m^(2)`