1 g charcoal adsorbs 100 ml of 0.5 M CH3CHOOH to form a monolayer. As a result molarity of acetic acid reduces to 0.49 M. What will be the surface area covered by each molecule of acetic acid? Given that surface area of charcoal =3.01×102m2/g.

1 g charcoal adsorbs 100 ml of 0.5 M CH3CHOOH to form a monolayer. As

1.	1 g charcoal adsorbs 100 ml of 0.5 M CH3CHOOH to form a monolayer. As a result molarity of acetic acid reduces to 0.49 M. What will be the surface area covered by each molecule of acetic acid? Given that surface area of charcoal =3.01×102m2/g.
Answer» 1 g charcoal adsorbs 100 ml of 0.5 M $C H_{3} C H O O H$ to form a monolayer. As a result molarity of acetic acid reduces to 0.49 M. What will be the surface area covered by each molecule of acetic acid? Given that surface area of charcoal $= 3.01 \times 10^{2} m^{2} / g$ .