In a solution of 100 mL 0.5 M acetic acid, one g of active charcoal is added, which adsorbs acetic acid. It is found that the concentration of acetic acid becomes 0.49 M if surface area of charcoal is `3.01 xx 10^(2) m^2`, calculate the area occupied by single acetic acid molecule on surface of charcoal.

Options

A. `2.5 xx 10^(-19) m^2`
B. `5 xx 10^(-19)m^2`
C. `0.5 xx 10^(-19) m^2`
D. `3.5 xx 10^(-19) m^2`

Answer

B

Explanation

The number of moles of acetic acid in 100 mL (before adding charcoal)=0.05 mol
The number of moles of acetic acid in 100 mL (after adding charcoal) = 0.049 mol
The number of moles of acetic acid adsorbed on the surface of charcoal=0.05-0.049=0.001 mol
The number of molecules of acetic acid adsorbed on the surface of charcoal `= 0.001 xx 6.02 xx 10^(23) = 6.02 xx 10^(20)`
Given that the surface area of charcoal `= 3.01 xx 10^2 m^2,` so the area occupied by single acetic acid molecule on the surface of charcoal is `(3 xx 10^2)/(6.02 xx 10^(20)) = 5 xx 10^(-19)m^2`