`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`.

`1 g` charcoal is placed in `100 mL` of `0.5 M CH_(3)COOH` to form an

1.	`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» Correct Answer - `5xx10^(19)m^(2)` 2. Initial millimoles of `CH_(3)COOH=100xx0.5=50` Final millimoles of `CH_(3)COOH=100xx0.49=49` Millimoles adsorbed `=50-49=1` Moles adsobed `=(1)/(1000)` Molecules adsorbed `=(1)/(1000)xx6.023xx10^(23)` `=6.023xx10^(20)` Area per molecule `=("Total area")/("Number of molecules")` `=(3.01xx10^(2))/(6.023xx10^(20))` `5xx10^(19)m^(2)`