Question

`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`.

Answer 1

Correct Answer - `(55.56 mol L^(-1))`
Initial millmol of `CH_(3)COOH = 100 xx 0.5 = 50`
millmol of `CH_(3)COOH` remaining after adsorption
`= 100 xx 0.49 = 49`
`rArr` millimol of `CH_(3)COOH` adsorbed `= 50 -49 = 1`
`rArr` number of molecules of `CH_(3)COOH` adsorbed
`= (1)/(1000) xx 6.023 xx 10^(23) = 6.023 xx 10^(20)`
`rArr` Area covered up by one molecule `= (3.01 xx 10^(2))/(6.02 xx 10^(20))`
`= 5 xx 10^(-19) m^(2)`