Question

1g of charcoal adsorbs 100ml of 0.5M CH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49M. what is the surface area of charcoal used by each molecule of acetic acid.? [surface area of charcoal =3.01*10²m²/g]​

Answer 1

Answer:

➡️ 100mL of 0.5M CH3COOH contains CH3COOH =0.05 mol.

➡️ after adsorption, CH3COOH remained =0.049mol

➡️ acetic acid adsorbed =0.001mol=6.02*10 ^20 molecules

➡️ surface area of charcoal adsorbed by each molecule =3.81*10²m²/6.02*10^20

=5*10^-19m²