1 gm of charcoal adsorbs 100 mL 0.7 M CH3COOH to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.59. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid in terms of 10-19. Surface area of charcoal = 3.01 × 102 m2/gm.

Question

1 gm of charcoal adsorbs 100 mL 0.7 M CH3COOH to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.59. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid in terms of 10-19. Surface area of charcoal = 3.01 × 102 m2/gm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Number of moles of acetic acid in 100 mL before adding charcoal = 0.07 Number of moles of acetic acid in 100 ml after adding charcoal = 0.059 Number of moles of acetic acid adsorbed on the surface of charcoal. Number of molecules of acetic acid adsorbed on the surface of charcoal = 0.011 × 6.02 × 1023 = 6.62 × 1021 Surface area of charcoal = 3.01 × 102 m2 (given) Area occupied by single acetic acid molecule on the surface of charcoal = (3.01 × 102/6.62 × 1021) = 0.45 × 10-19 m2

Q. 1gm of charcoal adsorbs 100mL0.7MCH3​COOH to form a monolayer, and thereby the molarity of CH3​COOH reduces to 0.59. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid in terms of 10−19. Surface area of charcoal =3.01×102m2/gm.