One gm of charcoal adsorbs CH 3 COOH from 100 ml 0.6 M CH 3 COOH aqueous solution to form a monolayer, and thereby the molarity of CH 3 COOH reduces to 0.58. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal =3.0 × 102 m 2 / gm.

Question

One gm of charcoal adsorbs CH 3 COOH from 100 ml 0.6 M CH 3 COOH aqueous solution to form a monolayer, and thereby the molarity of CH 3 COOH reduces to 0.58. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal =3.0 × 102 m 2 / gm. (A) 5 nm 2 (B) 0.5 nm 2 (C) 0.25 nm 2 (D) 2.5 nm 2

Tardigrade Admin · Accepted Answer

mm of CH 3 COOH adsorbed =2 number of molecule =2 × 10-3 × 6 × 1023 =12 × 1020 area adsorbed by each molecule =(3 × 106/12 × 1020) =0.25 × 10-14 cm 2 =25 × 10-16 cm 2=0.25 × 10-14 cm 2 =0.25 nm 2

$5 n m^{2}$

$0.5 n m^{2}$

$0.25 n m^{2}$

$2.5 n m^{2}$

Q. One gm of charcoal adsorbs CH3​COOH from 100ml0.6MCH3​COOH aqueous solution to form a monolayer, and thereby the molarity of CH3​COOH reduces to 0.58. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal =3.0×102m2/gm.

5nm2

0.5nm2

0.25nm2

2.5nm2

mm of CH3​COOH adsorbed =2 number of molecule =2×10−3×6×1023 =12×1020 area adsorbed by each molecule =12×10203×106​ =0.25×10−14cm2 =25×10−16cm2=0.25×10−14cm2 =0.25nm2

$5 n m^{2}$

$0.5 n m^{2}$

$0.25 n m^{2}$

$2.5 n m^{2}$