1 gm of charcoal adsorbs 100 ml 0 5 M CH3COOH to form a monolayer thereby the molarity of CH3COOH - Chemistry - Surface Chemistry

Question

1 gm of charcoal adsorbs 100 ml 0 5 M CH3COOH to form
a monolayer thereby the molarity of CH3COOH reduces to 0
49 Calculate the surface area of the charcoal adsorbed by each
molecule of acetic acid Surface area of charcoal= 3 01x
102 m2/gm

Arti Gujrati · Accepted Answer

Number of moles of acetic acid in given sample = concentration x
volume
= 0 5 mol L-1 x 0 1 L
= 0 05 moles
Number of moles of acetic acid left after adsorption = 0 49 mol
L-1 x 0 1 L
= 0 049 moles
Number of moles of acetic acid adsorb on surface = 0 05 - 0 049 = 0
001 mol
Number of molecules adsorbs on surface = 0 001 x 6 022 x
1023
= 6 022 x 1020
Surface Area occupied by each molecule = total surface area of
charcoal / Number of molecules of acetic acid
= 3 01 x 102 m2 g-1/ 6 022 x
1020
= 5 0 x 10-19 m2 g-1

1 gm of charcoal adsorbs 100 ml 0.5 M CH3COOH to form a monolayer thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal= 3.01x 102 m2/gm.

1 gm of charcoal adsorbs 100 ml 0.5 M CH₃COOH to form a monolayer thereby the molarity of CH₃COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal= 3.01x 10² m²/gm.