1 g charcoal adsorbs 100 mL of 0.5 M CH 3 COOH 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 × 102 m 2 g.

Question

1 g charcoal adsorbs 100 mL of 0.5 M CH 3 COOH 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 × 102 m 2 g. (A) 2.5 × 10-19 m 2 (B) 5.0 × 10-19 m 2 (C) 10-18 m 2 (D) 2.0 × 10-18 m 2

Tardigrade Admin · Accepted Answer

Acetic acid adsorbed by 1 g of charcoal 0.001 mole =6.02 × 1020 molecules. Surface area of 1 g of charcoal =3.01 × 102 m 2 ∴ Surface area of charcoal covered by each molecule. =(3.01 × 102 m 2) /(6.02 × 1020) =5 × 10-19 m 2

Q. $1 g$ charcoal adsorbs $100 m L$ of $0.5 M C H_{3} COO H$ 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 $= 3.01 \times 1 0^{2} m^{2} g$ .

$2.5 \times 1 0^{- 19} m^{2}$

$5.0 \times 1 0^{- 19} m^{2}$

$1 0^{- 18} m^{2}$

$2.0 \times 1 0^{- 18} m^{2}$

Acetic acid adsorbed by $1 g$ of charcoal
$0.001$ mole $= 6.02 \times 1 0^{20}$ molecules.
Surface area of $1 g$ of charcoal $= 3.01 \times 1 0^{2} m^{2}$
$∴$ Surface area of charcoal covered by each molecule.
$= (3.01 \times 1 0^{2} m^{2}) / (6.02 \times 1 0^{20})$
$= 5 \times 1 0^{- 19} m^{2}$

Q. 1g charcoal adsorbs 100mL of 0.5MCH3​COOH to form a monolayer. As a result molarity of acetic acid reduces to 0.49M. What will be the surface area covered by each molecule of acetic acid? Given that surface area of charcoal =3.01×102m2g.

2.5×10−19m2

5.0×10−19m2

10−18m2

2.0×10−18m2