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 - `5xx10^(-19) m^(2)`
`100 ml` of `0.5 CH_(3)COOH` contains `=0.05` mole
after adsorption, `CH_(3)COOH` presnet `=0.049` mole
acetic acid adsorbed by `1` gm charcoal `=0.05-0.049=0.01 "mole"=6.023xx10^(20)` molecule
surface area of `1` gm charcoal `=3.01xx10^(2)`
Surface area of charcoal adsorbed by each molecule `=3.01xx10^(2)//6.023xx10^(20)=5xx10^(-19) m^(2)`