`1 g` charcoal is placed in `100 mL` of `0.5 M CH_(3)COOH` to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of `CH_(3)COOH` reduces to `0.49`. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal `=3.01xx10^(2)m^(2)//g`.

Explanation

Milli-mole of acetic acid taken `=100xx0.5=50`
Milli-mole of acetic acid left after adsorption
`=100xx0.49=49`
`:.` Milli-mole of acetic acid absorbed `=50-49=1`
Number of acetic acid molecules absorbed
`=1xx10^(-3)xx6.023xx10^(23)=6.023xx10^(20)`
`:'` Absorption is monolayer and all these molecules are absorbed on total area of charcal `(3.01 xx 10^(2) m^(2))`
`:.` Area of 1 molecule of acetic acid `= (3.01 xx 10^(2))/(6.023 xx 10^(20))`
`= 5 xx 10^(-19) m^(2)`