One gram of charcoal adsorbs 100 mL of 0.5 `M CH_(3)COOH` to form a mono-layer and thereby the molarity of acetic acid is reduced to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface acid of charcoal `=3.01xx10^(2) m^(2)//gm`

Options

A. `2.5xx10^(-19) m^(2)`
B. `5.0xx10^(-19)m^(2)`
C. `10^(-18) m^(2)`
D. `2.0 xx10^(-18) m^(2)`

Answer

B

Explanation

Acetic aid adsorbed =0.5 -0.49 =0.01 M `:.` Acetic acid absorbed form 100 MI solution = 0.001 mole
Acetic acid adsorbed vy 1 g of charcoal = 0.001 mole = `6.02xx10^(20)` molecules
Surface area of 1 g of charcoa = `3.01 xx10^(2) m^(2)`
surface area if 1 g of charcol = `3.01 xx10^(2) m^(2)`
`:.` Surface area of charcoal covered by each molecule
`= (3.01xx10^(2)m^(2)) //(6.02xx10^(20))=5xx10^(-19) m^(2)`